1013. Battle Over Cities (25)

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最新推荐文章于 2022-01-29 13:29:31 发布

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本文介绍了一种算法，用于解决战争中城市被占领后，高速公路网络如何快速修复以保持其余城市间的连接。通过构建图模型并使用广度优先搜索（BFS），该算法能够高效计算出需要修复的高速公路数量。

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1013. Battle Over Cities (25)
时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue

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.
Sample Input
3 2 3
1 2
1 3
1 2 3
Sample Output
1
0
0

求抹去点之后的连通分量即可，结果为连通分量减一

#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <iostream>
#include <cstring>
#include <iomanip>
#include <string>
#include <cfloat>
#include <queue>
#include <map>

using namespace std;
const int MaxN = 1010;
vector<int>G[MaxN];
bool isInq[MaxN];
int N, C, num;

void BFS(int s)
{
    if (s == C)
    {
        --num;
        return;
    }
    queue<int>que; que.push(s); isInq[s] = true;

    while (que.size())
    {
        int u = que.front(); que.pop();
        for (int i = 0; i < G[u].size(); ++i)
        {
            int v = G[u][i];
            if (!isInq[v])
            {
                que.push(v);
                isInq[v] = true;
            }
        }
    }
}

int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG

    std::ios::sync_with_stdio(false);

    int M,K;
    cin >> N >> M >> K;
    for (int i = 0; i < M; ++i)
    {
        int u, v;
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }

    while (K--)
    {
        cin >> C;
        memset(isInq, 0, sizeof(isInq));
        isInq[C] = true;

        num = 0;
        BFS(1);
        for (int i = 1; i <= N; ++i)
        {
            if (!isInq[i])
            {
                BFS(i);
                ++num;
            }
        }

        cout << num << endl;
    }


    return 0;
}