1013 Battle Over Cities (25分)

最新推荐文章于 2022-11-22 19:07:26 发布

原创最新推荐文章于 2022-11-22 19:07:26 发布 · 59 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

126 篇文章

订阅专栏

本文探讨了在战争中保持城市间高速公路连接的重要性，提出了快速评估需修复高速公路数量的方法。当某一城市被敌方占领，所有进出该城市的高速路将关闭，文章提供了一种算法，用于即时计算为维持剩余城市连接所需的最小修复工作量。

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:

Sample Output:

1
0
0

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

void BFS(int u);
int n,m,k;    
int a[1000][1000] = {};
bool visit[1000] = {false};
int lost_city;
int main(){
    
    
    
    // for(int i = 0; i < 1000; i++){

    //     for (int j = 0; i < 1000; j++)
    //     {
    //         cout << a[i][j] << " ";
    //     }
    //     cout << endl;
    // }
        

        
    cin >> n >> m >> k;
    for(int i = 0; i < m; ++i){
        
        int c1,c2;
        cin >> c1 >> c2;

        a[c1][c2] = a[c2][c1] = 1;

    }

    // for (int i = 1; i <= n; i++)
    // {
        
    //     for (int j = 1; j <= n; j++)
    //     {
    //         cout << a[i][j] << " ";
    //     }
    //     cout << endl;
    // }
    



    for(int i = 0; i < k; ++i){

        cin >> lost_city;
        memset(visit, false, sizeof(visit));
        int count = 0;

        visit[lost_city] = true;
        
        
        for(int j = 1; j <= n; ++j){

            if(j != lost_city && !visit[j]){
                BFS(j);
                count++;
            }

        }
        cout << count - 1 << endl;


    }


   

    // for(int i = 0; i < v.size(); i++){
    //     int count = 0;
    //     for(int x = 1; x < n; x++){

    //         for(int y = x + 1; y <= n; y++){

    //             if(x != v[i] && y != v[i]){
    //                 if(!a[x][y])
    //                     count ++;
    //             }


    //         }

    //     }
    //     cout << count << endl;
    // }

    return 0;
}

void BFS(int u){

    queue<int> q;
    q.push(u);
    visit[u] = true;
    while (!q.empty())
    {
        int z = q.front();
        if(z == lost_city)
            return;
        q.pop();

        for(int i = 1; i <= n; ++i){

            if(a[z][i] && !visit[i] && i != lost_city){
                visit[i] = true;
                q.push(i);
                
            }

        }
    }
    


    
}