[PAT-甲级]1013.Battle Over Cities

最新推荐文章于 2020-07-20 21:26:05 发布

caicaiatnbu

最新推荐文章于 2020-07-20 21:26:05 发布

阅读量1.7k

点赞数

CC 4.0 BY-SA版权

文章标签：图遍历

本文链接：https://blog.youkuaiyun.com/caicaiatnbu/article/details/74911820

PAT甲集专栏收录该内容

25 篇文章

订阅专栏

本文介绍了一种算法，用于解决战争中城市被占领后如何快速确定需要修复的高速公路数量，以保持剩余城市的连接。通过遍历无向图确定新增连通块的数量，进而计算所需修复的高速公路数量。

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

解题思路：

给定一个无向图graph，当删除图中某个顶点时，并且与已删顶点相连的边也一并删除。求此时需添加几条边才可以使得整个图graph中所有顶点都是连通的。

对图graph中所有节点进行遍历，因为遍历过程总是访问到单个连通块，并将该连通块中所有节点的都标记为已访问。转换为求有多少个连通块。需要添加的边数，即为连通块数-1

代码如下：

#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;

vector<int> graph[1010];
bool visit[1010];

int curSearchPoint;

void DFS(int v)
{
  if(v == curSearchPoint)
    return ;
    
  visit[v] = true;
  for(int i = 0; i < graph[v].size(); i ++)
  {
    if(visit[graph[v][i]] == false)
      DFS(graph[v][i]);
  }
}


int main()
{
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for(int i = 0; i < m; i ++)
  {
    int a, b;
    scanf("%d%d", &a, &b);
    graph[a].push_back(b);
    graph[b].push_back(a);
  }

  for(int query=0; query < k; query ++)
  {
    scanf("%d", &curSearchPoint);
    memset(visit, false, sizeof(visit));
    
    int block = 0;
    for(int i = 1; i <= n; i ++)
    {
      if(i != curSearchPoint && visit[i] == false)
      {
        DFS(i);
        block ++;
      }
    }
    printf("%d\n", block-1);
  }
  return 0;
}