POJ 3041 Asteroids（二分匹配模板题）

最新推荐文章于 2019-04-11 23:30:14 发布

转载最新推荐文章于 2019-04-11 23:30:14 发布 · 72 阅读

本文探讨了如何使用二分图匹配算法解决Asteroids游戏中消除所有陨石所需的最少射击次数问题。通过建立陨石位置的行和列之间的二分图，我们能够找到最小点覆盖数，即最大匹配数，从而得出解决方案。

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Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10288		Accepted: 5556

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

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).

Source

USACO 2005 November Gold

常规建图方法。

模板

/*
poj 3041
行和列建立二分图
一个点上有东西则建立一条边。
题目就相当于求最小点覆盖数=最大二分匹配数
*/

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


/* **************************************************************************
//二分图匹配（匈牙利算法的DFS实现）
//初始化：g[][]两边顶点的划分情况
//建立g[i][j]表示i->j的有向边就可以了，是左边向右边的匹配
//g没有边相连则初始化为0
//uN是匹配左边的顶点数，vN是匹配右边的顶点数
//调用：res=hungary();输出最大匹配数
//优点：适用于稠密图，DFS找增广路，实现简洁易于理解
//时间复杂度:O(VE)
//***************************************************************************/
//顶点编号从0开始的
const int MAXN=510;
int uN,vN;//u,v数目
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];
bool dfs(int u)//从左边开始找增广路径
{
    int v;
    for(v=0;v<vN;v++)//这个顶点编号从0开始，若要从1开始需要修改
      if(g[u][v]&&!used[v])
      {
          used[v]=true;
          if(linker[v]==-1||dfs(linker[v]))
          {//找增广路，反向
              linker[v]=u;
              return true;
          }
      }
    return false;//这个不要忘了，经常忘记这句
}
int hungary()
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=0;u<uN;u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}
//******************************************************************************/



int main()
{
     int k;
     int n;
     int u,v;
     while(scanf("%d%d",&n,&k)!=EOF)
     {
         uN=vN=n;
         memset(g,0,sizeof(g));
         while(k--)
         {
             scanf("%d%d",&u,&v);
             u--;
             v--;
             g[u][v]=1;
         }
         printf("%d\n",hungary());
     }
     return 0;
}