ACM: 匈牙利算法 图论题 poj 3041 …

                                                               Asteroids

Description

Bessie wants to navigate her spaceship through a dangerousasteroid field in the shape of an N x N grid (1 <= N<= 500). The grid contains K asteroids (1<= K <= 10,000), which areconveniently located at the lattice points of the grid.

Fortunately, Bessie has a powerful weapon that can vaporize all theasteroids in any given row or column of the grid with a singleshot.This weapon is quite expensive, so she wishes to use itsparingly.Given the location of all the asteroids in the field,find the minimum number of shots Bessie needs to fire to eliminateall of the asteroids.

Input

* Line 1: Two integers N and K, separated by a singlespace.
* Lines 2..K+1: Each line contains two space-separated integers Rand C (1 <= R, C <= N) denoting therow and column coordinates of an asteroid, respectively.

Output

* Line 1: The integer representing the minimum number of timesBessie must shoot.

Sample Input

3 4

1 1

1 3

2 2

3 2

Sample Output

2

Hint

INPUT DETAILS:
The following diagram represents the data, where "X" is an asteroidand "." is empty space:
X.X
.X.
.X.

OUTPUT DETAILS:
Bessie may fire across row 1 to destroy the asteroids at (1,1) and(1,3), and then she may fire down column 2 to destroy the asteroidsat (2,2) and (3,2).

 

题意: Bessie 要消灭grid中所有的障碍, 她每发射一次可以摧毁一行 or 一列 的障碍.

         现在要用最少的发射次数消灭全部的障碍.

 

解题思路:

               1. 将行列分成两个集合, 现在要用最少的行的点,匹配列的点.

               2. 把问题抽象为二分图, 最大匹配的问题.

               3. 匈牙利算法水之.

 

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 505

int n, m;
int graph[MAX][MAX];
bool vis[MAX];
int state[MAX];

void read_graph()
{
    int u,v;
   memset(graph,0,sizeof(graph));
    for(int i =1; i <= m; ++i)
    {
      scanf("%d%d",&u,&v);
      graph[u][v] = 1;
    }
}

bool find(int u)
{
    for(int v =1; v <= n; ++v)
    {
      if(graph[u][v] > 0&& !vis[v])
      {
         vis[v] = true;
         if(state[v] == -1 || find(state[v]))
         {
            state[v] = u;
            return true;
         }
      }
    }
    returnfalse;
}

int Edmonds()
{
    int result =0;
   memset(state,-1,sizeof(state));
    for(int u =1; u <= n; ++u)
    {
      memset(vis,false,sizeof(vis));
      if(find(u))
         result++;
    }
    returnresult;
}

int main()
{
//   freopen("input.txt","r",stdin);
   while(scanf("%d%d",&n,&m) != EOF)
    {
      read_graph();
      printf("%d\n",Edmonds());
    }
    return0;
}