PAT Advanced 1013 Battle Over Cities（25）

题目描述

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 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

算法标签：DFS

题目大意：

给出n个城市，m条邻边，k次询问。
每次询问给出一个被敌人攻占的城市，求失去该城市后的连通块，答案即是连通块的数量-1。

#include<iostream>
#include<vector>
using namespace std;

int n, m, k;
bool st[1010];
int sign; // 被敌人攻占的城市
vector<int> v[1010];

void dfs(int u)
{
    st[u] = true;
    for(auto x : v[u])
        if(!st[x] && x != sign) // 没有被遍历过 且 非敌人攻占城市
            dfs(x);
}

int main()
{
    cin >> n >> m >> k;
    for(int i = 0 ; i < m ; i ++)
    {
        int x, y;
        scanf("%d %d", &x, &y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    
    while(k --)
    {
        cin >> sign;
        for(int i = 1 ; i <= n ; i ++)  st[i] = false;
        
        int j = 0; // 连通子图数量
        for(int i = 1 ; i <= n ; i ++)
            if(!st[i] && i != sign)
            {
                j ++;
                dfs(i);
            }
        cout << j - 1 << endl;
    }
    
    return 0;
}