前言

常用技巧

日期问题

具体到星期几【可使用windows计算机器】

package ch01;import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Arrays;
import java.util.Calendar;
import java.util.Date;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;public class Main {public static void main(String[] args) {// 年/月/日的，有采用月/日/年的，还有采用日/月/年的000Scanner cin = new Scanner(System.in);Calendar c = Calendar.getInstance();c.set(2020, 10,9);//月份从0开始System.out.println(c.get(Calendar.DAY_OF_WEEK) - 1);}}

【Windows计算器实现日期计算】【日期之差】【日期对应星期几】

… 日期格式化 - parse

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;public class Main {public static void main(String[] args) {// 年/月/日的，有采用月/日/年的，还有采用日/月/年的000Scanner cin = new Scanner(System.in);String a[] = cin.next().split("/");String t[] = new String[3];t[0] = ((Integer.valueOf(a[0]) >= 60) ? "19" + a[0] : "20" + a[0]) + "-" + a[1] + "-" + a[2];t[1] = ((Integer.valueOf(a[2]) >= 60) ? "19" + a[2] : "20" + a[2]) + "-" + a[0] + "-" + a[1];t[2] = ((Integer.valueOf(a[2]) >= 60) ? "19" + a[2] : "20" + a[2]) + "-" + a[1] + "-" + a[0];SimpleDateFormat sf = new SimpleDateFormat("yyyy-MM-dd");sf.setLenient(false);Set<String> s = new TreeSet<String>();for (String T : t) {try {sf.parse(T);} catch (ParseException e) {continue;}s.add(T);}for(String T:s)System.out.println(T);}
}

Java排序

Java优先队列重载

PriorityQueue<Node> pq = new PriorityQueue<>(new Comparator<Node>() {@Overridepublic int compare(Node o1, Node o2) {// TODO Auto-generated method stubreturn o1.costT - o2.costT;}});

交叉排序【内部类实现排序】【熟记】

交叉排序

import java.math.BigInteger;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
public class Main {public static void main(String[] args) {Scanner cin = new Scanner(System.in);Integer a[];Integer n = cin.nextInt();a = new Integer[n + 5];int l1 = cin.nextInt();int r1 = cin.nextInt();int l2 = cin.nextInt();int r2 = cin.nextInt();for(int i = 1;i <= n;i++)a[i] = cin.nextInt();Arrays.sort(a, l1, r1 + 1);Arrays.sort(a, l2, r2 + 1, new Comparator<Integer>(){@Overridepublic int compare(Integer o1, Integer o2) {// TODO Auto-generated methodreturn o2 - o1;}});for(int i = 1;i <= n;i++)if(i == 1)System.out.print(a[i]);else System.out.print(" " + a[i]);System.out.println();}
}

内部类排序【动态数组,排序自定义,foreach输出】

成绩排序

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;public class Main {static class node{int c; int m; int e; int s;int sum;String name;node(String name,int c, int m, int e, int s){this.name = name;this.c = c;this.m = m;this.e = e;this.s = s;this.sum = c + m + e + s;}}public static void main(String[] args) {Scanner cin = new Scanner(System.in);ArrayList<node>ar = new ArrayList<node>();int n = cin.nextInt();for(int i = 1;i <= n;i++){String name = cin.next();int c = cin.nextInt();int m = cin.nextInt();int e = cin.nextInt();int s = cin.nextInt();node nd = new node(name, c, m, e, s);ar.add(nd);}ar.sort(new Comparator<node>() {@Overridepublic int compare(node o1, node o2) {// TODO Auto-generated method stubif(o1.c == o2.c)return o1.name.compareTo(o2.name);    else return o2.c - o1.c;}});int cnt = 0;for(node T:ar) {cnt++;if(cnt == 1)System.out.print(T.name);else System.out.print(" " + T.name);if(cnt == 4)break;}System.out.println();ar.sort(new Comparator<node>() {@Overridepublic int compare(node o1, node o2) {// TODO Auto-generated method stubif(o1.m == o2.m)return o1.name.compareTo(o2.name);   else return o2.m - o1.m;}});cnt = 0;for(node T:ar) {cnt++;if(cnt == 1)System.out.print(T.name);else System.out.print(" " + T.name);if(cnt == 4)break;}System.out.println();ar.sort(new Comparator<node>() {@Overridepublic int compare(node o1, node o2) {// TODO Auto-generated method stubif(o1.e == o2.e)return o1.name.compareTo(o2.name);   else return o2.e - o1.e;}});cnt = 0;for(node T:ar) {cnt++;if(cnt == 1)System.out.print(T.name);else System.out.print(" " + T.name);if(cnt == 4)break;}System.out.println();ar.sort(new Comparator<node>() {@Overridepublic int compare(node o1, node o2) {// TODO Auto-generated method stubif(o1.s == o2.s)return o1.name.compareTo(o2.name);   else return o2.s - o1.s;}});cnt = 0;for(node T:ar) {cnt++;if(cnt == 1)System.out.print(T.name);else System.out.print(" " + T.name);if(cnt == 4)break;}System.out.println();ar.sort(new Comparator<node>() {@Overridepublic int compare(node o1, node o2) {// TODO Auto-generated method stubif(o1.sum == o2.sum)return o1.name.compareTo(o2.name);   else return o2.sum - o1.sum;}});cnt = 0;for(node T:ar) {cnt++;if(cnt == 1)System.out.print(T.name);else System.out.print(" " + T.name);if(cnt == 4)break;}System.out.println();}
}

正则表达式（用于处理分割字符串）

详解链接
题目链接

class Solution {public int numDifferentIntegers(String word) {if(word.length() == 0) return word.length();String newStr = word.replaceAll("[a-z]+", " ");//代替a到z的全部元素String[] s = newStr.split(" ");HashMap<String, Integer> map = new HashMap<>();for (String s1 : s) {if(s1.equals("")){continue;}String s2 = s1.replaceAll("^(0+)", "");//匹配从0开头所有元素0map.put(s2, map.getOrDefault(s2, 0)+1);//如果map种没有s2这个key，则用0作为它的value}return map.size();}
}

图论

最小生成树

import java.util.*;public class Main {static class Node{int u, v, w;Node(int u, int v, int w){this.u = u;this.v = v;this.w = w;}}static Node nd[];static Vector<Integer> ar[];static int f[];static int find(int x){if(x == f[x])return x;return f[x] = find(f[x]);}static void setUnion(int x, int y){x = find(x);y = find(y);f[x] = y;}public static void main(String[] args) {Scanner cin = new Scanner(System.in);int n = cin.nextInt();n = n * (n - 1) / 2;f = new int[n + 5];nd = new Node[n + 5];for(int i = 0;i < n;i++){int u = cin.nextInt();int v = cin.nextInt();int w = cin.nextInt();int f = cin.nextInt();if(f == 1)w = 0;nd[i] = new Node(u, v, w);}Arrays.sort(nd, 0, n, new Comparator<Node>() {@Overridepublic int compare(Node o1, Node o2) {return o1.w - o2.w;}});int cost = 0;for(int i = 1;i <= n;i++)f[i] = i;for(int i = 0;i < n;i++){int x = find(nd[i].u);int y = find(nd[i].v);if(x == y)continue;//都在一棵树上，passcost += nd[i].w;f[x] = y;}System.out.println(cost);}
}

最短路

floyd模板

算法分析

代码：

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner cin = new Scanner(System.in);int n = cin.nextInt();int m = cin.nextInt();int x = cin.nextInt();int dis[][] = new int[n + 5][n + 5];for(int i = 1;i <= n;i++)for(int j = 1;j <= n;j++)if(i == j)dis[i][j] = 0;elsedis[i][j] = 99999999;for(int i = 1;i <= m;i++) {int u = cin.nextInt();int v = cin.nextInt();int w = cin.nextInt();dis[u][v] = Math.min(w, dis[u][v]);}for(int k = 1;k <= n;k++)for(int i = 1;i <= n;i++)for(int j = 1;j <= n;j++)dis[i][j] = Math.min(dis[i][j], dis[i][k] + dis[k][j]); while(x--!=0) {int u = cin.nextInt();int v = cin.nextInt();if(dis[u][v] >= 99999999 / 2)System.out.println("impossible");else System.out.println(dis[u][v]);}}
}

牛的旅行 floyd(多源最短路)

题目链接

import java.util.Scanner;public class Main {static double INF = 0x3f3f3f3f;static double len(double x1, double y1, double x2, double y2){return Math.sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));  }public static void main(String[] args) {Scanner cin = new Scanner(System.in);int n = cin.nextInt();double zb[][] = new double[n + 5][2];char map[][] = new char[n + 5][n + 5];double dis[][] = new double[n + 5][n + 5];double maxDis[] = new double[n + 5];for(int i = 1;i <= n;i++){zb[i][0] = cin.nextInt();zb[i][1] = cin.nextInt();}for(int i = 1;i <= n;i++){String s = cin.next();for(int j = 1;j <= s.length();j++)map[i][j] = s.charAt(j - 1);}for(int i = 1;i <= n;i++)for(int j = 1;j <= n;j++)if(i == j)dis[i][j] = 0;else dis[i][j] = INF;for(int i = 1;i <= n;i++)for(int j = 1;j <= n;j++)if(map[i][j] == '1')dis[i][j] = len(zb[i][0], zb[i][1], zb[j][0], zb[j][1]);for(int k = 1;k <= n;k++)for(int i = 1;i <= n;i++)for(int j = 1;j <= n;j++)dis[i][j] = Math.min(dis[i][j], dis[i][k] + dis[k][j]);double ans1 = 0;for(int i = 1;i <= n;i++)for(int j = 1;j <= n;j++)if(dis[i][j] != INF){maxDis[i] = Math.max(maxDis[i], dis[i][j]);ans1 = Math.max(ans1, maxDis[i]);}double ans = INF;for(int i = 1;i <= n;i++)for(int j = 1;j <= n;j++)if(dis[i][j] > INF / 2){ans = Math.min(ans, maxDis[i] + len(zb[i][0], zb[i][1], zb[j][0], zb[j][1]) + maxDis[j]);}System.out.printf("%.6f",Math.max(ans, ans1));}
}

数论

欧拉筛法【O(nlogn)级别快速求质数】

链接
长度为11的等差素数列，其公差最小值是多少？

package ch01;public class isPrime {static int []prime = new int[100005];static boolean []vis = new boolean[100005];static int cnt = 0, count = 0;static void sieve(int n){cnt = 0;for(int i = 2; i <= n; i++){if(!vis[i]) //不是目前找到的素数的倍数 是素数prime[cnt++] = i; //找到素数for(int j = 0; j < cnt && i * prime[j] <= n; j++){vis[i * prime[j]] = true; //找到的素数的倍数不访问if(i % prime[j] == 0) break; //关键！！！！}}}public static void main(String[] args) {sieve(100000);  for(int i = 1;i <= 1000;i++) {for(int j = 0;j < cnt;j++) {int sum = prime[j];int k = 1;while(true) {count++;if(sum > 100000 || vis[sum] || k == 11) {//不是素数breakbreak;}sum += i;k++;}if(k == 11) {System.out.println(i);System.out.println(count);System.exit(0);}}}}
}

辗转相除法求最大公约数【熟记】

static int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);
}

动态规划

二维动态规划

题目链接

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Vector;class Main {public static void main(String[] args) {Scanner cin = new Scanner(System.in);int n = cin.nextInt();int sum = cin.nextInt();int dp[][] = new int[n + 15][sum + 15];dp[1][1] = 1;dp[1][2] = 1;dp[1][3] = 1;dp[1][4] = 1;dp[1][5] = 1;dp[1][6] = 1;for(int i = 1;i <= n;i++)for(int j = 1;j <= sum;j++) for(int k = 1;k <= 6 && j > k;k++)dp[i][j] = (dp[i][j] + dp[i - 1][j - k])%998244353; System.out.println(dp[n][sum]);    }
}

数据结构

HashMap 嵌套 Set

题目链接

class Solution {public int[] findingUsersActiveMinutes(int[][] logs, int k) {int ans[] = new int[k];HashMap<Integer, Integer>ansp = new HashMap<>();HashSet<Integer>sst = new HashSet<>();HashMap<Integer, HashSet<Integer>>mp = new HashMap<>();for(int i = 0;i < logs.length;i++){int id = logs[i][0];//idint t = logs[i][1];//时间sst.add(id);if(mp.containsKey(id)){if(!mp.get(id).contains(t)){mp.get(id).add(t);// mp.put(id, st);ansp.put(id, ansp.get(id) + 1);// System.out.println(ansp.get(id));}}else{HashSet<Integer>st = new HashSet<>();st.add(t);mp.put(id, st);ansp.put(id,1);}// System.out.println(ansp.get(id));}for(Integer t:sst)ans[ansp.get(t) - 1]++;return ans;}
}

队列

栈

并查集

使用并查集判断连通块数量【模板】

题目连接

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Vector;class Main {static int n, m;static int fa[] = new int[20005];static void makeSet(int size) {for (int i = 1; i <= size; i++)fa[i] = i; // i就在它本身的集合里return;}static int find(int x) {if (x != fa[x]) // x不是自身的父亲，即x不是该集合的代表fa[x] = find(fa[x]); // 查找x的祖先直到找到代表,于是顺手路径压缩return fa[x];}static void unionSet(int x, int y) {// x 与 y 所在家族合并x = find(x);y = find(y);fa[x] = y;  // 把 x 的祖先变成 y 的祖先的儿子}public static void main(String[] args) {Scanner cin = new Scanner(System.in);n = cin.nextInt();m = cin.nextInt();makeSet(n);for (int i = 1; i <= m; i++) {int u = cin.nextInt();int v = cin.nextInt();unionSet(u, v);}int cnt = 0;for(int i = 1;i <= n;i++)if(fa[i] == i)cnt++;System.out.println(cnt);}
}

import java.util.*;public class Main {static class Node{String name;int num;Node(String name, int num){this.name = name;this.num = num;}}static Vector<Integer> ar[];static int f[];static int find(int x){if(x == f[x])return x;return f[x] = find(f[x]);}static void setUnion(int x, int y){x = find(x);y = find(y);f[x] = y;}public static void main(String[] args) {Scanner cin = new Scanner(System.in);int n = cin.nextInt();int m = cin.nextInt();ar = new Vector[n + 5];f = new int[n + 5];int flag = 0;for(int i = 0;i < n + 5;i++)ar[i] = new Vector<>();for(int i = 1;i <=n;i++ )f[i] = i;for(int i = 0;i < m;i++){int u = cin.nextInt();int v = cin.nextInt();setUnion(u,v);ar[u].add(v);ar[v].add(u);}for(int i = 1;i <= n;i++){if(ar[i].size() % 2 == 1)flag = 1;}int cnt = 0;for(int i = 1;i <= n;i++)if(i == f[i])cnt++;if(cnt == 1 && flag == 0) System.out.println("1");else System.out.println("0");}
}

优先队列

标题：第几个幸运数到x星球旅行的游客都被发给一个整数，作为游客编号。
x星的国王有个怪癖，他只喜欢数字3,5和7。
国王规定，游客的编号如果只含有因子：3,5,7,就可以获得一份奖品。我们来看前10个幸运数字是：
3 5 7 9 15 21 25 27 35 45
因而第11个幸运数字是：49小明领到了一个幸运数字 59084709587505，他去领奖的时候，人家要求他准确地说出这是第几个幸运数字，否则领不到奖品。请你帮小明计算一下，59084709587505是第几个幸运数字。需要提交的是一个整数，请不要填写任何多余内容。

import java.math.BigInteger;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.TreeSet;public class Test2 {public static void main(String[] args) {PriorityQueue<Long> q = new PriorityQueue<Long>();TreeSet<Long> st = new TreeSet<Long>();q.add((long) 1);int cnt = 0;while (!q.isEmpty()) {long x = q.poll();System.out.println(x);if(("" + x).equals("59084709587505"))break;if (!st.contains(x * 3)) {q.add(x * 3);st.add(x * 3);cnt++;}if (!st.contains(x * 5)) {q.add(x * 5);st.add(x * 5);cnt++;}if (!st.contains(x * 7)) {q.add(x * 7);st.add(x * 7);cnt++;}}cnt = 0;for(Long T:st) {cnt++;if((""+T).equals("59084709587505")) {System.out.println(cnt);break;}}}
}

二分

寻找第一个大于目标数的数

package ch01;import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Arrays;
import java.util.Date;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;public class Main {public static void main(String[] args) {// 年/月/日的，有采用月/日/年的，还有采用日/月/年的000Scanner cin = new Scanner(System.in);int n = cin.nextInt();long vis[] = new long[100005];int a[][] = new int[3][n + 5];for (int i = 0; i < n; i++) {a[0][i] = cin.nextInt();}for (int i = 0; i < n; i++) {a[1][i] = cin.nextInt();}for (int i = 0; i < n; i++) {a[2][i] = cin.nextInt();}Arrays.sort(a[0], 0, n);Arrays.sort(a[1], 0, n);Arrays.sort(a[2], 0, n);long res = 0;int ff = 0;long max = 0;for (int i = 0; i < n; i++) {int l = 0, r = n;while (l <= r) {int mid = (l + r) >>> 1;if (a[0][i] < a[1][mid]) {r = mid - 1;} elsel = mid + 1;}int pos = a[1][l];
//              System.out.println(pos);if (l >= n)break;if (l != 0) {if (vis[a[1][l]] != 0)res += max - vis[a[1][l - 1]];} elseres += max;
//          System.out.println(res + " " + a[1][l] + " " + vis[a[1][l - 1]]);
//              System.out.println(pos);
//              System.out.println(vis[a[1][l - 1]]);int flag = l;if (ff == 0) {while (true) {l = 0;r = n;ff = 1;while (l <= r) {int mid = (l + r) >>> 1;if (a[1][flag] < a[2][mid]) {r = mid - 1;} elsel = mid + 1;}
//              System.out.println(a[2][l] + " sd");if (l >= n)break;res += (long) (n - l);if (flag == 0)vis[a[1][flag]] = n - l;elsevis[a[1][flag]] = res;flag++;max = res;if (flag == n)break;}}}System.out.println(res);}}

搜索

dfs

2020年JavaB组国赛 E:玩具蛇

思路： 经典dfs,搜出每一种情况。

public class Main {static int map[][] = new int[20][20];static int cnt;static void dfs(int x, int y, int num) {if(x < 1 || y < 1 || x > 4 || y > 4)return ;if(map[x][y] != 0)return ; map[x][y] = num;if(num == 16) {cnt++;for(int i = 1;i <= 4;i++) {for(int j = 1;j <= 4;j++)System.out.print(map[i][j] + " ");System.out.println();}System.out.println();map[x][y] = 0;return ;}dfs(x + 1, y, num + 1);dfs(x - 1, y, num + 1);dfs(x, y + 1, num + 1);dfs(x, y - 1, num + 1);map[x][y] = 0;}public static void main(String[] args) {for(int i = 1;i <= 4;i++)for(int j = 1;j <= 4;j++)dfs(i, j, 1);System.out.println(cnt);}
}

判断是否能表示成3的幂数和

题目链接

class Solution {static int f;void dfs(long sum, int pos, int index){// System.out.println(sum + " " + pos);if(f == 1 || sum > pos)return ;if(sum == pos){f = 1;return ;}for(int i = index;i <= 15;i++){dfs(sum + (int)Math.pow(3, i), pos, i + 1);}}public boolean checkPowersOfThree(int n) {f = 0;dfs(0, n, 0);    if(f == 1)return true;return false;}
}

搜索枚举各类情况

题目链接

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Vector;class Main {static String s;static int k;static int min;static void dfs(String s, int z, int res) {if(res < 0)return ;//System.out.println(res);if(res == 0) {System.out.println(k + "=" +s.substring(0, s.length() - 1));return;}for(int i = z;i < k;i++)dfs(s + i + '+', i, res - i);}public static void main(String[] args) {Scanner cin = new Scanner(System.in);k = cin.nextInt();dfs("", 1, k);}
}

题目链接

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Vector;class Main {static String s;static int k;static int min;static void dfs(int l, int r, int cs, int res) {if (cs == k - 1) {//          System.out.println("-->" + (res + Integer.valueOf(s.substring(r, s.length()))));min = Math.min(res + Integer.valueOf(s.substring(r, s.length())), min);return;}for (int i = r + 1; i < s.length(); i++) {//          System.out.println(s.substring(r, i));dfs(r, i, cs + 1, res + Integer.valueOf(s.substring(r, i)));}}public static void main(String[] args) {Scanner cin = new Scanner(System.in);s = cin.next();k = cin.nextInt();min = 9999999;dfs(0, 0, 0, 0);System.out.println(min);}
}

类似八皇后

题目连接

import java.util.Arrays;
import java.util.Scanner;
import java.util.Vector;class Main {static int n, m;static int vis[]  = new int[10 + 5];static char a[][] = new char[10 + 5][10 + 5];;static int cnt;static void dfs(int r, int sp) {if (m == sp) {cnt++;return;}if(r == n + 1)return ;for (int i = 0; i < n; i++) {if (a[r][i] == '#' && vis[i] == 0) {vis[i] = 1;dfs(r + 1, sp + 1);vis[i] = 0;}}dfs(r + 1, sp);}public static void main(String[] args) {Scanner cin = new Scanner(System.in);while(true) {n = cin.nextInt();m = cin.nextInt();if(n == -1 && m == -1)break;for (int i = 0; i < n; i++) {String s = cin.next();for (int j = 0; j < n; j++) {a[i][j] = s.charAt(j);
//              System.out.println(s);}}cnt = 0;dfs(0, 0);System.out.println(cnt);}}
}

判断联通块数量

题目连接

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Vector;class Main {static int n, m;static ArrayList<Integer> ar[] = new ArrayList[20005];static int vis[] = new int[20005];static void dfs(int u) {for (int i = 0; i < ar[u].size(); i++) {if (vis[ar[u].get(i)] == 0) {vis[ar[u].get(i)] = 1;dfs(ar[u].get(i));}}}public static void main(String[] args) {Scanner cin = new Scanner(System.in);n = cin.nextInt();m = cin.nextInt();for (int i = 0; i < 20005; i++)ar[i] = new ArrayList<>();for (int i = 0; i < m; i++) {int u = cin.nextInt();int v = cin.nextInt();ar[u].add(v);ar[v].add(u);}int cnt = 0;for (int i = 1; i <= n; i++)if (vis[i] != 1) {cnt++;vis[i] = 1;dfs(i);}System.out.println(cnt);}
}

bfs

迷宫【bfs模板】

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;public class Main {// date1 = "2019-06-29", date2 = "2019-06-30"static class Node{int x;int y;int step;Node(int x, int y, int step){this.x = x;this.y = y;this.step = step;}}public static void main(String[] args) {Scanner cin = new Scanner(System.in);int dir[][] = {{1, 0},{-1, 0},{0, 1},{0, -1}};int vis[][];char a[][];int tran[][][];int n = cin.nextInt();int m = cin.nextInt();a = new char[n + 5][m + 5];vis = new int[n + 5][m + 5];tran = new int[n + 5][m + 5][2];for (int i = 1; i <= n; i++) {String s = cin.next();for (int j = 1; j <= m; j++) {a[i][j] = s.charAt(j - 1);}}int qx = cin.nextInt();for(int i = 0;i < qx;i++) {int x = cin.nextInt();int y = cin.nextInt();tran[x][y][0] = cin.nextInt();tran[x][y][1] = cin.nextInt();}int tx = cin.nextInt();int ty = cin.nextInt();Queue<Node> q = new LinkedList<Node>();q.add(new Node(1, 1, 0));while(!q.isEmpty()) {Node node = q.poll();if(node.x == tx && node.y == ty) {System.out.println(node.step);System.exit(0);         }if(tran[node.x][node.y][0] != 0) {int dx = tran[node.x][node.y][0];int dy = tran[node.x][node.y][1];if(dx == tx && dy == ty) {System.out.println(node.step);System.exit(0);         }if(a[dx][dy] == '*') {System.out.println("No solution");System.exit(0);}else {if(vis[dx][dy] == 1)continue;vis[dx][dy] = 1;q.add(new Node(dx, dy, node.step));
//                  continue;}continue;}for(int i = 0; i < 4;i++) {int nx = node.x + dir[i][0];int ny = node.y + dir[i][1];if(nx <= 0 || nx > n || ny <= 0 || ny > m || vis[nx][ny] == 1 || a[nx][ny] == '*')continue;vis[nx][ny] = 1;q.add(new Node(nx, ny, node.step + 1));}}System.out.println("No solution");}
}

背包

01背包

01背包问题

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Vector;class Main {public static void main(String[] args) {Scanner cin = new Scanner(System.in);int N = cin.nextInt();int V = cin.nextInt();int v[] = new int[1005];int w[] = new int[1005];int dp[] = new int[1005];for(int i = 1;i <= N;i++) {v[i] = cin.nextInt();w[i] = cin.nextInt();}for(int i = 1;i <= N;i++)for(int j = V;j >= v[i];j--)dp[j] = Math.max(dp[j], dp[j - v[i]] + w[i]);System.out.println(dp[V]);}
}

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