1013 Battle Over Cities （25 分)

2019-03-09

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**y1-cit**y2 and cit**y1-cit**y3. Then if cit**y1 is occupied by the enemy, we must have 1 highway repaired, that is the highway cit**y2-cit**y3.

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
2
3

1
0
0

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

有n座城市，其中一座被敌军攻占，需修多少路可以将剩余城市连接起来。

分析

利用dfs求图的连通分量数即可。被攻占的城市visit初始化为1。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <cstring>
#include <cmath>

using namespace std;
int n;
bool visit[1010]={};
int graph[1010][1010]={};

void dfs(int node) {
    visit[node] = true;
    for(int i = 1; i <= n; i++) {
        if(visit[i] == false && graph[node][i] == 1)
            dfs(i);
    }
}
int main() {
    std::ios::sync_with_stdio(false);
    int m,k,c1,c2,t,cnt=0;
    cin>>n>>m>>k;

    for(int i=0;i<m;i++)
    {
        cin>>c1>>c2;
        graph[c1][c2]=graph[c2][c1]=1;
    }

    for(int i=0;i<k;i++){
        cin>>t;
        cnt=0;
        fill(visit,visit+1010,0);
        visit[t]=1;
        for(int j=1;j<=n;j++){
            if(visit[j]==0){
                dfs(j);
                cnt++;
            }
        }
        cout<<cnt-1<<endl;
    }

}