ACM: 匈牙利算法 图论题 poj 3041 …

最新推荐文章于 2019-04-07 20:06:41 发布

Smile_Benson

最新推荐文章于 2019-04-07 20:06:41 发布

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本文介绍了一种通过最小发射次数清除网格中所有障碍的方法。利用二分图的最大匹配原理，采用匈牙利算法求解最少射击次数问题。

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Asteroids

Description

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

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

Input

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

Output

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

Sample Input

3 4

1 1

1 3

2 2

3 2

Sample Output

Hint

INPUT DETAILS:
The following diagram represents the data, where "X" is an asteroid and "." 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 asteroids at (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;
           }
       }
   }
   return false;
}

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++;
   }
   return result;
}

int main()
{
//   freopen("input.txt","r",stdin);
   while(scanf("%d %d",&n,&m) != EOF)
   {
       read_graph();
       printf("%d\n",Edmonds());
   }
   return 0;
}