1013 Battle Over Cities

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本文探讨了在战争背景下，确保城市间高速公路连通性的快速修复算法。通过图的遍历找强连通子集数量，确定需要修复的高速公路数量，以保持剩余城市的连接。使用DFS深度优先搜索算法进行求解。

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 city1-city2 and city1-city3. Then if city1 is occupied by the enemy, we must have 1 highway repaired, that is the highway city2-city3.
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:

Sample Output:

1
0
0

题解：

乍一看是最小生成树问题，但仔细分析会发现是利用图的遍历来找强连通子集数量的问题，即有 $n$ 个连通子集，至少需要 $n - 1$ 条边使他们相连，可以使用 DFS 求解

#include <iostream>
#include <cstdlib>
#include <cstring>
using namespace std;

int N,M,K,V,W;
bool road[1001][1001] = {0};
bool visit[1001];

void DFS(int i)
{
    visit[i] = true;
    for(int j=0;j<N;j++)
        if(road[i][j] == 1 && visit[j] == 0)
            DFS(j);
}

int main()
{
    cin>>N>>M>>K;
    for(int i=0;i<M;i++){
        cin>>V>>W;
        road[V-1][W-1] = road[W-1][V-1] = 1;
    }

    for(int i=0;i<K;i++){
        int cont = 0;
        memset(visit,0,sizeof(visit));
        cin>>V;
        visit[--V]=true;
        for(int j=0;j<N;j++){
            if(!visit[j]){		// 连通集判断
                DFS(j);
                cont++;
            }
        }
        cout<<cont-1<<endl;
    }
}