题目

1013 Battle Over Cities (25分)

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.

For example, if we have 3 cities and 2 highways connecting $cit{y}_1$-$cit{y}_{}2$ and $cit{y}_{}1$​​ -$cit{y}_3$​ . Then if $cit{y}_1$​ is occupied by the enemy, we must have 1 highway repaired, that is the highway $cit{y}_2$ - $cit{y}_{}3$ .

Input Specification:

Each input file contains one test case. Each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. Then M lines follow, each describes a highway by 2 integers, which are the numbers of the cities the highway connects. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.

Output Specification:

For each of the $K$ cities, output in a line the number of highways need to be repaired if that city is lost.

Sample Input:

3 2 3 1 2 1 3 1 2 3

Sample Output:

1 0 0

题目大意

给出N个城市，M条道路，每条路连通两个城市，然后给出K个沦陷的城市，对其中每个沦陷的城市，求出当它沦陷的时候，为了使剩余的城市连通，最少需要多少条路。

思路

对应无向图求连通分量个数问题，n个连通分量则需要n-1条路，借助DFS来求连通分量个数。

代码

#include<bits/stdc++.h> using namespace std; int N, M, K, ans; vector<int> edge[1010]; bool visit[1010] = {false}; void dfs(int st){ visit[st] = true; for(int i=0; i<edge[st].size(); i++){ if(!visit[edge[st][i]]) dfs(edge[st][i]); } } int main(int argc, const char * argv[]) { int c1, c2; cin>>N>>M>>K; for(int i=0; i<M; i++){ cin>>c1>>c2; edge[c1].push_back(c2); edge[c2].push_back(c1); } for(int i=0; i<K; i++){ cin>>c1; ans = 0; fill(visit, visit+N+1, false); visit[c1] = true; for(int j=1; j<=N; j++){ if(!visit[j]){ dfs(j); ans++; } } printf("%d\n", ans-1); } return 0; }