PAT A1013 Battle Over Cities （25 分）——图遍历，联通块个数

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 Knumbers, 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

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
const int maxn = ;
int g[maxn][maxn],l[maxn][maxn];
int n, m, k, lost;
bool vis[maxn];
void dfs(int i) {
    if (vis[i] == false) {
        vis[i] = true;
        for (int j = ; j <= n; j++) {
            if (l[i][j] != ) {
                dfs(j);
            }
        }
    }
}
int dfsTrave() {
    int count = ;
    fill(vis, vis + maxn, false);
    vis[lost] = true;
    for (int i = ; i <= n; i++) {
        if (vis[i] == false) {
            dfs(i);
            count++;
        }
    }
    return count;
}
int main() {
    fill(g[], g[] + maxn * maxn, );
    scanf("%d %d %d", &n,&m,&k);
    for (int i = ; i < m; i++) {
        int c1, c2;
        scanf("%d %d\n",&c1, &c2);
        g[c1][c2] = ;
        g[c2][c1] = ;
    }
    for (int i = ; i < k; i++) {
        scanf("%d", &lost);
        //copy(l[0], l[0] + maxn * maxn, g);
        memcpy(l, g, maxn*maxn * sizeof(int));
        for (int j = ; j <= n; j++) {
            if (l[lost][j] != ) {
                l[lost][j] = ;
                l[j][lost] = ;
            }
        }
        /*for (int i = 0; i <= 5+n; i++) {
            for (int j = 0; j <= 5+n; j++) {
                printf("%d ", l[i][j]);
            }
            printf("\n");
        }*/
        int res = dfsTrave()-;
        printf("%d\n", res);
    }
    system("pause");
}

注意点：仔细分析题目后发现其实考的就是一个图有几个联通块，用dfs遍历一遍就可以了。二维数组的复制要用到 string.h 头文件下的 memcpy 函数，不知道为什么copy函数一直报错