1013 Battle Over Cities

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题目

1013 Battle Over Cities (25)（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 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

Each input file contains one test case. Each case starts with a line containing 3 numbers N (&lt1000), 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

1
2
3
4
3 2 3
1 2
1 3
1 2 3

Sample Output

1
2
3
1
0
0

想法

图算法，连通分量，需要修建的就是连通分量的数量-1

连通分量怎么求呢，早忘了，尴尬
只是数出来而已
BFS/DFS应该都可以
准备一个mark数组，visit就标记，碰到没有标记的点，分量+1

写了一个很臃肿的版本

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//
// Created by Zhao Xiaodong on 2018/7/30.
//
#include <iostream>
#include <string>
#include <vector>
#include <climits>
#include <algorithm>
using namespace std;
void dfs(vector<vector<int>> &m, int k, vector<bool> &visited) {
    if (visited[k]) return;
    visited[k] = true;
    for (int i = 1; i < m[k].size(); i++) {
        if (m[k][i]) {
            dfs(m, i, visited);
        }
    }
}
int main() {
    int N, M, K;
    cin >> N >> M >> K;
    vector<vector<int>> m(N + 1, vector<int>(N + 1, 0));
    for (int i = 0; i < M; i++) {
        int a, b;
        cin >> a >> b;
        m[a][b] = m[b][a] = 1;
    }
    for (int i = 0; i < K; i++) {
        int t;
        cin >> t;
        vector<vector<int>> tmp = m;
        for (int j = 1; j <= N; j++) {
            tmp[t][j] = tmp[j][t] = 0;
        }
        vector<bool> visited(N + 1, false);
        visited[t] = true;
        int res = 0;
        for (int k = 1; k <= N; k++) {
            if (visited[k])
                continue;
            res++;
            dfs(tmp, k, visited);
        }
        cout << res - 1 << endl;
    }
    return 0;
}

竟然超时了？？？

F**k，换scanf就过了。。。

答案

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//
// Created by Zhao Xiaodong on 2018/7/30.
//
#include <iostream>
#include <string>
#include <vector>
#include <climits>
#include <algorithm>
using namespace std;
void dfs(vector<vector<int>> &m, int k, vector<bool> &visited) {
    if (visited[k]) return;
    visited[k] = true;
    for (int i = 1; i < m[k].size(); i++) {
        if (m[k][i]) {
            dfs(m, i, visited);
        }
    }
}
int main() {
    int N, M, K;
    cin >> N >> M >> K;
    vector<vector<int>> m(N + 1, vector<int>(N + 1, 0));
    for (int i = 0; i < M; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        m[a][b] = m[b][a] = 1;
    }
    for (int i = 0; i < K; i++) {
        int t;
        cin >> t;
        vector<vector<int>> tmp = m;
        for (int j = 1; j <= N; j++) {
            tmp[t][j] = tmp[j][t] = 0;
        }
        vector<bool> visited(N + 1, false);
        visited[t] = true;
        int res = 0;
        for (int k = 1; k <= N; k++) {
            if (visited[k])
                continue;
            res++;
            dfs(tmp, k, visited);
        }
        cout << res - 1 << endl;
    }
    return 0;
}

总结

做PAT，能用scanf、printf，就不要用cin、cout，真的是太坑了。。。

能用全局变量就用全局变量，OJ嘛，管那么多干嘛，所有的都全局，省的传参麻烦

还是没有做OJ的经验啊