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 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:

3 2 3
1 2
1 3
1 2 3

Sample Output:

1
0
0

题意：给个图，给出k次查询，每次查询给出一个结点，要求判断去掉这个结点之后图是否是连通图，输出使图连通需要添加的边的条数

思路：建图，对于每次查询，dfs算法求图中极大连通子图的个数，输出的结果是子图个数-1

代码如下 ：

#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,k;
bool visited[1005]={false};
bool G[1005][1005]={false};
void dfs(int v,int city){
    visited[v]=true;
    for(int i=1;i<=n;i++){
        if(G[v][i]==true){
            if(i!=city&&visited[i]==false){
                dfs(i,city);
            }    
        }
    }
}
void dfsTraverse(int v){
    fill(visited,visited+n+1,false);
    int count=-1;
    for(int i=1;i<=n;i++){
        if(i!=v&&visited[i]==false){
            count++;
            dfs(i,v);
        }
    }
    printf("%d\n",count);
}

int main(){
    scanf("%d%d%d",&n,&m,&k);
    int a,b;
    for(int i=1;i<=m;i++){
        scanf("%d%d",&a,&b);
        G[a][b]=G[b][a]=true;
    }
    int city;
    for(int i=0;i<k;i++){
        scanf("%d",&city);
        dfsTraverse(city);
    }
    return 0;
}

 

标签：city,city1,pat,int,highways,cities,repaired,1013
来源： https://www.cnblogs.com/foodie-nils/p/13337001.html