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

解析：题意很简单，在一个连通图中如果去掉一个节点和该节点的所有edge，那么需要修复多少条edge才能保持连通图。该题的思路之一就是使用DFS查找图的连通性，然后连通图的个数减1就是所需要修复的edge数。

Code:

/*************************************************************************
    > File Name: 1013.cpp
    > Author:
    > Mail:
    > Created Time: 2015年12月15日 星期二 20时27分37秒
 ************************************************************************/
#include<iostream>
#include<fstream>
#include<cstring>

using namespace std;
const int size=1010;
int edge[size][size];
int visited[size];
int K, M, N;
void dfs(int i)
{
    visited[i]=1;
    for(int j=1; j<=N; j++)
    {
        if(!visited[j] && edge[i][j])
            dfs(j);
    }
}

int main()
{
    memset(edge, 0, sizeof(edge));
    cin>>N>>M>>K;
    for(int i=0,a, b; i<M; i++)
    {
        cin>>a>>b;
        edge[a][b] = edge[b][a] = 1;
    }
    int tmp;
    while(K--)
    {
        cin>>tmp;
        memset(visited, 0, sizeof(visited));
        int ans=0;
        visited[tmp]=1;
        for(int i=1; i<=N; i++)
        {
            if(!visited[i])
            {
                dfs(i);
                ans++;
            }
        }
        cout<<ans-1<<endl;
    }
    return 0;
}

还有BFS方法，不过貌似最后一个测试点过不了，这里也贴上代码。

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int size=1010;
int edge[size][size];
int K, M, N;
int visited[size];

int main()
{
    memset(edge,0,sizeof(edge));

    cin>>N>>M>>K;
    for(int i=0,a,b; i<M; i++)
    {
        cin>>a>>b;
        edge[a][b]=edge[b][a]=1;
    }

    int tmp;
    queue<int>que;
    while(K--)
    {
        cin>>tmp;
        memset(visited,0,sizeof(visited));

        int ans=0;
        visited[tmp]=1;
        for(int i=1; i<=N; i++)
        {
            if(!visited[i])
            {
                que.push(i);
                while(!que.empty())
                {
                    int j=que.front();
                    que.pop();
                    visited[j]=1;
                    for(int k=1; k<=N; k++)
                        if(!visited[k]&&edge[j][k])
                                que.push(k);
                }
                ans++;
            }
        }
        cout<<ans-1<<endl;
    }
    return 0;
}