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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.
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.
3 2 3
1 2
1 3
1 2 3
Sample Output:
1
0
0
题目大意：
给出节点和边，求如果其中某个节点及与之相关的边无效后，其他节点组成连通图还需要的最少边数。
解题思路：
并查集。使用DFS求去掉某点和它的边后，一共有几个独立的子图。
一个包含N的节点的连通图的最小边数是N-1。
源代码：
#include<iostream>
#include<stdlib.h>
using namespace std;
const int inf = 99999999;
int n, m, k, e[1001][1001], book[1001];
void dfs(int i, int c)
{
book[i] = c;
for (int j = 1; j <= n; j++)
{
if (e[i][j] == 1 && book[j] == 0)
dfs(j, c);
}
return;
}
int main()
{
int i, j,a,b;
scanf("%d %d %d", &n, &m, &k);
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++)
if (i == j)e[i][j] = 0;
else e[i][j] = inf;
for (i = 0; i < m; i++)
{
scanf("%d %d", &a, &b);
e[a][b] = e[b][a] = 1;
}
for (i = 0; i < k; i++)
{
for (j = 1; j <= n; j++)book[j] = 0;
scanf("%d", &a);
book[a] = a;
int cnt = 0;
for (j = 1; j <= n; j++)
{
if (book[j] == 0)
{
cnt++;
dfs(j, cnt);
}
}
printf("%d\n", cnt - 1);
}
system("pause");
return 0;
}
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