九度题目1325：Battle Over Cities_至少修几条道路java-优快云博客

本文链接：https://blog.youkuaiyun.com/wuyexiaobailong/article/details/71171791

本文介绍了一种算法，用于解决战争中城市被占领后，为保持剩余城市间交通连接所需的高速公路修复数量问题。通过使用深度优先搜索算法，快速计算出需要修复的高速公路数量。

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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.

输入：

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.

输出：

For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.

样例输入：

样例输出：

1 0

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这个等价于求一个图的连通分量有多少个，有K个连通分量需要修K-1条路，明显用深度优先搜索法，上代码：


#include<iostream>
#include<string.h>
using namespace std;
bool v[1001][1001];//只需要bool值表示通不通就可以了
bool used[1001];
int n,m;
bool dfs(int j)
{
    int k;
    for(k=1;k<=n;k++)
        if((!used[k])&&v[j][k])
        {
            used[k]=true;
            dfs(k);
        }
    return true;
}
int maxl(int a)
{
    used[a]=true;//将第a个城市设置成已经访问过，等效于被敌人占领了
    int j,k=0;
    for(j=1;j<=n;j++)//从1开始深度遍历
    {
        if(!used[j])
        {
            used[j]=true;
            dfs(j);
            k++;//每遍历一次表示找到了一个连通分量
        }，统计下来
    }
    return k;
}
int main()
{
    int k,j,i;
    int a,b;
    while(cin>>n>>m>>k)
    {
        memset(v,false,sizeof(v));
        for(j=0;j<m;j++)
        {
            cin>>a>>b;
            v[a][b]=true;
            v[b][a]=true;
        }
        for(j=0;j<k;j++)
        {
            memset(used,false,sizeof(used));
            cin>>a;
            cout<<maxl(a)-1<<endl;
        }
    }
    return 0;
}
/**************************************************************
    Problem: 1325
    User: 午夜小白龙
    Language: C++
    Result: Accepted
    Time:460 ms
    Memory:2500 kb
****************************************************************/