The 15th Chinese Northeast Collegiate Programming Contest K.City

题目：

思路：

代码：

题目：

原题：

Lucida occupies n cities connected by m undirected roads, and each road has a strength ki. The enemy will attack to destroy these roads. When the enemy launches an attack with damage x, all roads with strength less than x will be destroyed.

Now Lucida has Q questions to ask you, how many pairs of cities are reachable to each other if the enemy launches an attack with damage pi. City x and city y are reachable, which means that there is a path from x to y, and every road's strength in that path is greater than or equal to pi.

简述：

给定两两城市之间的道路长度和一系列伤害值。强度大于等于伤害值的道路不会被破坏。问对于每一个伤害值，有多少对可以互相到达的城市。

思路：

存储：用结构体存道路的起始和权值，以及询问的伤害和序号。重载分别按权值和伤害降序排序。

算法：并查集，维护集合大小。

操作：

将道路按权值降序排序（为了先连接可连接的城市），询问按伤害降序排序（伤害小的集合大小可以在伤害大的集合基础上累加）。对于每一个伤害值，累加新连通的两个集合的城市数，将可合并且未合并的集合合并，并更新集合大小。

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;typedef long long ll;
const int N = 2e5 + 5;
int p[N];      //记祖先和集合的大小
ll size[N], ans[N];struct node {int x, y, k;bool operator < (const node &rd)const {  //按路的权值降序(先搭能搭的)return k > rd.k;}
} road[N];struct query {int id, p;bool operator < (const query &qy)const {  //按伤害的值降序(先搭条件严苛的)return p > qy.p;}
} q[N];int find(int x) {                              //并查集return x == p[x] ? x : find(p[x]);
}int main() {int T;cin >> T;while (T --) {int n, m, Q;cin >> n >> m >> Q;memset(ans, 0, sizeof ans);for (int i = 1; i <= n ; ++ i) {  //初始化p[i] = i;size[i] = 1;}for (int i = 0; i < m; ++ i) {int x, y, k;cin >> x >> y >> k;road[i] = {x, y, k};}for (int i = 0; i < Q; ++ i) {cin >> q[i].p;q[i].id = i;}sort(road, road + m);          //排序sort(q, q + Q);ll sum = 0;int j = 0;for (int i = 0;  i < Q; ++ i) {int damage = q[i].p;while (j < m && road[j].k >= damage) {  //能搭int u = find(road[j].x);int v = find(road[j].y);if (u != v) { //未搭sum += (ll)size[u] * size[v];  //为两个集合的城市搭桥p[u] = v;                      //合并顶点size[v] += size[u];            //合并大小}++ j;}ans[q[i].id] = sum;}for (int i = 0; i < Q; ++ i) cout << ans[i] << endl;}return 0;
}

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The 15th Chinese Northeast Collegiate Programming Contest K.City

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代码：

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