1013. Battle Over Cities (25)

最新推荐文章于 2022-01-29 13:29:31 发布

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最新推荐文章于 2022-01-29 13:29:31 发布

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本文介绍了一个算法挑战，旨在解决战争背景下城市间高速公路网络的连通性问题。当某个城市被敌方占领时，所有进出该城市的道路将被封锁。为保持其余城市间的连接，需要立即确定哪些额外的道路需要修复。通过输入剩余城市数量、可用道路及关注城市的数据，文章提供了快速判断所需修复道路数量的方法。

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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

Sample Output

1
0
0

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
bool Map[1000][1000];
bool vis[1000];
int Mindis;
int n,m,k;
void dfs(int x)
{
   for(int i =1; i <= n ;i++)
   {
      if(vis[i] == 0 && Map[x][i] != 0)
      {
          vis[i] = 1;
          dfs(i);
      }
   }
}
int main()
{
   scanf("%d%d%d", &n, &m, &k);
   int u,v;
   memset(Map, 0, sizeof(Map));
   for(int i = 0; i < m; i++)
   {
      scanf("%d%d", &u,&v);
      Map[u][v] = Map[v][u] = 1;
   }
   int x;
   while(k--)
   {
       scanf("%d", &x);
       memset(vis, 0, sizeof(vis));
       vis[x] = 1;
       int cnt = 0;
       for(int i = 1; i <= n; i++)
       {
          if(vis[i] == 0)
          {
             dfs(i);
             cnt++;
          }
       }
       printf("%d\n",cnt - 1);

   }
}