1013. Battle Over Cities (25)-优快云博客

本文针对1013.BattleOverCities问题，介绍了两种解决方法：深度优先搜索（DFS）和并查集算法。该问题关注城市间高速公路连接，在某城市被占领后，需修复多少道路以保持其余城市连通。通过实例演示了算法的具体实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

1
0

为什么我每次都会把简单的问题搞复杂，这道题的模型是求有几个连通的子集n，则需要修n-1条路。

看了几位大佬的文章，好多是用并查集算法做的，http://blog.youkuaiyun.com/dm_vincent/article/details/7655764这篇从并

查集出发，谈论了好多，值得收藏。http://blog.youkuaiyun.com/liujian20150808/article/details/50848646这篇很生动啊

我本能想到的是用DFS算法，还可以记录路径。

方法一：DFS

#include<stdio.h>  
#include<string.h>  
  
#define max 1001  
int edge[max][max];  
int visited[max];  
int query[max];  
int N; // the total number of cities  
int M; // the number of remaining highways  
int K; // the number of cities to be checked  
  
void dfs(int t)  
{  
    visited[t] = 1;  
    int i;  
    for(i=1; i<=N; i++)  
    {  
        if(!visited[i] && edge[i][t] == 1)  
            dfs(i);  
    }  
}  
  
  
  
int main()  
{     
    int i,j;  
    int a,b;      
    scanf("%d%d%d",&N,&M,&K);  
    for(i=0; i<M; i++)  
    {  
        scanf("%d%d",&a,&b);  
        edge[a][b] = 1;  
        edge[b][a] = 1;  
    }  
    int temp;  
    int num;  
    for(i=0; i<K; i++)  
    {  
        num = 0;  
        scanf("%d",&temp);  
        memset(visited,0,sizeof(visited)); //数组的整体初始化，memset在<string.h>,<iostream>,<mem.h>中都有。

        visited[temp] = 1;  
        for(j=1; j<=N; j++)  
        {  
            if(visited[j] == 0)  
            {                 
                dfs(j);  
                num ++;  
            }  
        }  
        printf("%d\n",num-1);  
    }  
  
}

方法二：并查集

#include <cstdio>  
int Tree[1000];  
int findRoot(int x){  
    if (Tree[x] == -1) return x;  
    else{  
        int tmp = findRoot(Tree[x]);  
        Tree[x] = tmp;  
        return tmp;  
    }  
}  
struct w{  
    int x;  
    int y;  
}way[500000];  
int main(){  
    int n, m, k;  
    scanf("%d%d%d", &n, &m, &k);  
    for (int i = 0; i < m; i++){  
        scanf("%d%d", &way[i].x, &way[i].y);  
    }  
    while (k--){  
        for (int i = 1; i <= n; i++)  
            Tree[i] = -1;  
        int c;  
        scanf("%d", &c);  
        for (int i = 0; i < m; i++){  
            if (way[i].x != c && way[i].y != c){  
                int rx = findRoot(way[i].x);  
                int ry = findRoot(way[i].y);  
                if (rx != ry){  
                    Tree[rx] = ry;  
                }  
            }  
        }  
        int group = 0;  
        for (int i = 1; i <= n; i++){  
            if (Tree[i] == -1)  
                group++;  
        }  
        printf("%d\n", group-2);  
    }  
    return 0;  
}

注意并查集的表示在数组Tree中。