PAT (Advanced Level) 1013 Battle Over Cities (25分)

战时城市连接修复算法

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在战争中，保持城市间高速公路的连通性至关重要。若某城市被敌军占领，所有进出该城市的高速路将关闭。为了即时了解是否需要修复其他道路以保持剩余城市的连接，本文提出一种算法，通过地图上的城市和剩余高速公路信息，快速计算出若特定城市失守，需要修复的高速公路数量。

PAT (Advanced Level) 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 $city_1$ - $city_2$ and $c i t y 1$ - $city_3$ . 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

#include <bits/stdc++.h>
using namespace std;
vector <int> graph[1005];
int vis[1005], x;
void dfs(int u){
    if (u == x) return;
    for (auto v : graph[u]){
        if (vis[v] || v == x) continue;
        vis[v] = 1;
        dfs(v);
    }
}
int n, m, k;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> m >> k;
    for (int i = 0; i < m; i++){
        int u, v;
        cin >> u >> v;
        graph[u].push_back(v);
        graph[v].push_back(u);
    }

    for (int i = 0; i < k; i++){
        memset(vis, 0, sizeof(vis));
        cin >> x;
        int cnt = 0;
        for (int j = 1; j <= n; j++){
            if (!vis[j] && j != x){
                dfs(j);
                cnt++;
            }
        }
        cout << cnt - 1<< "\n";
    }
}