K - Treasure Exploration（匈牙利+Floyd）

最新推荐文章于 2020-03-02 17:04:41 发布

原创最新推荐文章于 2020-03-02 17:04:41 发布 · 366 阅读

0 ·

CC 4.0 BY-SA版权

ACM训练+实训+大学编程练习专栏收录该内容

362 篇文章

订阅专栏

本文介绍了一个用于优化火星探测任务中机器人部署数量的算法。通过构建一个由节点和单向路径组成的图模型，利用匈牙利算法寻找最小数量的机器人来覆盖所有探索区域，解决了EUC公司面临的财务和技术挑战。

Have you ever read any book about treasure exploration? Have you ever see any film about treasure exploration? Have you ever explored treasure? If you never have such experiences, you would never know what fun treasure exploring brings to you.
Recently, a company named EUC (Exploring the Unknown Company) plan to explore an unknown place on Mars, which is considered full of treasure. For fast development of technology and bad environment for human beings, EUC sends some robots to explore the treasure.
To make it easy, we use a graph, which is formed by N points (these N points are numbered from 1 to N), to represent the places to be explored. And some points are connected by one-way road, which means that, through the road, a robot can only move from one end to the other end, but cannot move back. For some unknown reasons, there is no circle in this graph. The robots can be sent to any point from Earth by rockets. After landing, the robot can visit some points through the roads, and it can choose some points, which are on its roads, to explore. You should notice that the roads of two different robots may contain some same point.
For financial reason, EUC wants to use minimal number of robots to explore all the points on Mars.
As an ICPCer, who has excellent programming skill, can your help EUC?
Input
The input will consist of several test cases. For each test case, two integers N (1 <= N <= 500) and M (0 <= M <= 5000) are given in the first line, indicating the number of points and the number of one-way roads in the graph respectively. Each of the following M lines contains two different integers A and B, indicating there is a one-way from A to B (0 < A, B <= N). The input is terminated by a single line with two zeros.
Output
For each test of the input, print a line containing the least robots needed.
Sample Input

Sample Output

1
1
2

要看好能够匹配的地方。。。。。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
const int MAXN = 567;
int uN, vN;
int linker[MAXN];
bool used[MAXN];
int g[MAXN][MAXN];
bool dfs(int u)
{
  for(int v = 0;v < vN; v++)
  {
    if(g[u][v] && !used[v])
    {
       used[v] = true;
       if(linker[v]==-1||dfs(linker[v]))
       {
         linker[v] = u;
         return true;
       }
    }
  }
  return false;
}
int hungry()
{
  int res = 0;
  memset(linker, -1, sizeof(linker));
  for(int u=0;u<uN;u++)
  {
    memset(used, false, sizeof(used));
    if(dfs(u))
    res++;
  }
  return res;
}
void Floyd()//补充可以匹配的点
{
  for(int i=0;i<uN;i++)
  {
    for(int j=0;j<uN;j++)
    {
      if(g[i][j]==0)
      {
        for(int k=0;k<uN;k++)
        {
          if(g[i][k]&&g[k][j])
          {
            g[i][j] = 1;
            break;
          }
        }
      }
    }
  }
}
int main()
{
   int n, m, a, b;
   while(~scanf("%d %d", &n, &m)&&(n||m))
   {
      memset(g, 0, sizeof(g));
      for(int i=1;i<=m;i++)
      {
        scanf("%d %d", &a, &b);
        g[a-1][b-1] = 1;
      }
      uN = vN = n;
      Floyd();
      printf("%d\n", n - hungry());
   }
   return 0;
}