PAT1013: Battle Over Cities

Posted 2020-10-09 0kk470

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

 

思路

这道题的本质问题其实就是求去掉某一点后的图有几个连通子图。那么需要重修的道路就是——连通子图数-1（保证余下节点全连通）。所以只要在去掉一个点后dfs下这个图确定连通子图个数就行。那么：

1.用一个数组表示点是否被遍历。

2.对于一个被入侵的城市city，只需要在dfs之前将这个城市标记为已遍历就行，这样dfs相当于只遍历去掉了这个点的图，即模拟了city从图中被去掉的情况。

代码

#include<iostream>
#include<vector>
using namespace std;

vector<vector<int>> graph(1000,vector<int>(1000,-1));

void dfs(const int city,const int num,vector<bool>& visits)
{
   visits[city] = true;
   for(int i = 1;i <= num;i++)
   {
       if(city == i)
        continue;
       if(graph[city][i] == 1 && !visits[i])
       {
           dfs(i,num,visits);
       }
   }
}

int main()
{
   int N,M,K;
   while(cin >> N >> M >> K)
   {
      while(M--)
      {
          int a,b;
          cin >> a >> b;
          graph[a][b] = graph[b][a] = 1;
      }

      while(K--)
      {
          int countRoad = 0;
          vector<bool> visits(N + 1,false);
          int city;
          cin >> city;
          visits[city] = true;
          for(int i = 1;i <= N;i++)
          {
              if(!visits[i])
              {
                  countRoad++;
                  dfs(i,N,visits);
              }
          }
          cout << countRoad - 1 << endl;
      }
   }
}