hdu5971 Wrestling Match（染色法判二分） 2016ACM/ICPC亚洲区大连站

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5971

Wrestling Match

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1100 Accepted Submission(s): 420

Problem Description

Nowadays, at least one wrestling match is held every year in our country. There are a lot of people in the game is "good player”, the rest is "bad player”. Now, Xiao Ming is referee of the wrestling match and he has a list of the matches in his hand. At the same time, he knows some people are good players,some are bad players. He believes that every game is a battle between the good and the bad player. Now he wants to know whether all the people can be divided into "good player" and "bad player".

Input

Input contains multiple sets of data.For each set of data,there are four numbers in the first line:N (1 ≤ N≤ 1000)、M(1 ≤M ≤ 10000)、X,Y(X+Y≤N ),in order to show the number of players(numbered 1toN ),the number of matches,the number of known "good players" and the number of known "bad players".In the next M lines,Each line has two numbersa, b(a≠b) ,said there is a game between a and b .The next line has X different numbers.Each number is known as a "good player" number.The last line contains Y different numbers.Each number represents a known "bad player" number.Data guarantees there will not be a player number is a good player and also a bad player.

Output

If all the people can be divided into "good players" and "bad players”, output "YES", otherwise output "NO".

Sample Input

5 4 0 0 1 3 1 4 3 5 4 5 5 4 1 0 1 3 1 4 3 5 4 5 2

Sample Output

NO YES

题意：给出n个人，m场比赛，x个人为good player，y个人为bad player，每场比赛两个人分分别为good和bad，问good和bad是否会冲突

思路：染色法判二分，先根据已知的人判断，再遍历一遍n个点找到还未判断的，wa了n次，忘记vector初始化的，，，整个人都不好了

代码：

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<cmath>
#include<algorithm>
#include<string>
#include<string.h>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<functional>
#include<map>
using namespace std;
const int INF = (int)1e9;
const int maxn = 25000 + 10;
typedef long long LL;//视情况定类型  int n, m, x, y;
vector<int>G[maxn];
int color[maxn];void init() {memset(color, 0, sizeof(color));for (int i = 0; i < maxn; i++)G[i].clear();
}int dfs(int s) {for (int i = 0; i < G[s].size(); i++) {int v = G[s][i];if (color[v] == 0) {color[v] = -color[s];if (!dfs(v))return 0;}else if (color[s] == color[v])return 0;}return 1;
}int main() {while (scanf("%d %d %d %d", &n, &m, &x, &y) != EOF) {init();int a, b;for (int i = 0; i < m; i++) {scanf("%d %d", &a, &b);G[a].push_back(b);G[b].push_back(a);}for (int i = 0; i < x; i++) {scanf("%d", &a);color[a] = 1;}for (int i = 0; i < y; i++) {scanf("%d", &b);color[b] = -1;}if ((x == 0 && y == 0) || n==1) {printf("NO\n");continue;}int ans = 1;for (int i = 1; i <= n; i++) {if (color[i]!=0 && dfs(i) == 0) {ans = 0;break;}}for (int i = 1; i <= n; i++) {if (color[i] == 0) {color[i] = 1;if (dfs(i) == 0) {ans = 0;break;}}}if (ans == 1)printf("YES\n");else printf("NO\n");}system("pause");return 0;
}

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hdu5971 Wrestling Match（染色法判二分） 2016ACM/ICPC亚洲区大连站

Wrestling Match

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