pat

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

一开始题目看错了，以为编号是乱序的，所以想到用set存一下出现过的索引号。后来一直有一个点过不了，仔细读题，发现编号是$1~N$，如果用set的话，中间有一些独立的连通分量，没有被发现。

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#include <iostream>
#include <vector>
using namespace std;
typedef vector<int> vi;
vector<vi> Map(1005, vi());
vi (1005, 0);
int N, M, K, a, b;
void dfs(int u){
    vst[u] = 1;
    for(int i = 0;i < (int)Map[u].size();i++){
        int v = Map[u][i];
        if(vst[v] == 0)
            dfs(v);
    }
}
int main(){
    scanf("%d %d %d", &N, &M, &K);
    for(int i = 0;i < M;i++){
        scanf("%d %d", &a, &b);
        Map[a].push_back(b);
        Map[b].push_back(a);
    }
    for(int i = 0;i < K;i++){
        fill(vst.begin(), vst.end(), 0);
        int cnt = 0;
        scanf("%d", &a);
        vst[a] = 1;
        for(int j = 1;j <= N;j++){
            if(vst[j] == 0){
                dfs(j);
                cnt++;
            }
        }
        printf("%dn", cnt-1);
    }
    return 0;
}