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

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

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 31 21 31 2 3Sample Output
100
標簽：圖論，DFS
思路：算出除去某個點后的強連通個數(shù)，強連通數(shù)-1就是要修的路的個數(shù)
#include<iostream>#include <iomanip>#include<vector>#include<set>#include<map>#include<string>#include<algorithm>using namespace std;int edge[1000][1000] = { 0 };bool isVisted[1000] = { false };int concern[1000];void DFS(int n,int node,int cur){    isVisted[cur] = true;    int i;    for ( i = 1; i <= n; i++)    {        if (!isVisted[i] && edge[cur][i] && i != node){  //剪掉node相關的點            DFS(n, node, i);        }    }}int numofDFS(int n,int node){    int num = 0,i;    isVisted[node] = true;    for ( i = 1; i <= n; i++)    {        if (!isVisted[i]){            DFS(n,node,i);            num++;        }    }    return num;  }int main(){    int n, m,k,i,j;    cin >> n >> m>>k;    int c1, c2;    for ( i = 0; i < m; i++)    {        cin >> c1 >> c2;        edge[c1][c2] = 1;        edge[c2][c1] = 1;    }    for ( i = 0; i < k; i++)    {        cin >> concern[i];    }    for ( i = 0; i < k; i++)    {        cout << numofDFS(n, concern[i])-1 << endl;        for ( j = 1; j <=n; j++)        {            isVisted[j] = false;  //重置標志數(shù)組        }    }}