PAT 1013 Battle Over Cities (25 分)

• 题目：

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 city​1​​-city​2​​ and city​1​​-city​3​​. Then if city​1​​ is occupied by the enemy, we must have 1 highway repaired, that is the highway city​2​​-city​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
  给定一些城市节点，城市间的高速公路可以看成为图的边，问去掉一个节点后需要增加几条边使得所有城市联通
• 解题方法
  只用求去掉一个节点后，剩下的图中有多少个最大联通图，所需要增加的边数就是最大联通图的数目减一
• 代码实现
  利用DFS遍历图即可。

//算法思想，计算最大联通分量，则需要修的高速公路数为联通分量减一
#include <cstdio>
#include <cstring>
const int MAX = 1001;
int N, M, K;
int G[MAX][MAX] = {0};            //没有通路则为0
int visit[MAX] = {false};
void DFS(int i, int node)
{   if(i == node)                         //不用真的删除节点，只用遇到这个节点返回即可
        return;
    visit[i] = true;
    for(int j = 1; j <= N; j++){
        if(G[i][j] == 1 && visit[j] == false){       //如果两节点之间存在边，且为访问
            DFS(j, node);
        }
    }
}
int travel(int curnode){
    int num = 0;
    for(int i = 1; i <= N; i++){
        if(visit[i] ==false && i != curnode)
        {
            num++;
            DFS(i, curnode);
        }
    }
    return num - 1;
}
int main()
{
    scanf("%d%d%d", &N, &M, &K);
    for(int i = 0; i < M; i++){
        int node1, node2;
        scanf("%d%d", &node1, &node2);
        G[node1][node2] = 1;
        G[node2][node1] = 1;                 //两节点之间存在边
    }
    for(int i = 0; i < K; i++){
        int node;
        scanf("%d", &node);
        memset(visit, false, sizeof(visit));     //初始化visit数组为false
        int ans = travel(node);
        printf("%d\n", ans);
    }
    return 0;
}