POJ P3041 Asteroids 题解

POJ P3041 Asteroids 题解

POJ P3041

题目

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.

输入

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.

输出

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

样例

input
3 4
1 1
1 3
2 2
3 2

output
2

样例解释

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

题目给出一个矩阵，上面有敌人，每个子弹可以打出一横行或者一竖行，问最少用多少子弹消灭都有敌人，如：
X.X
.X.
.X.
x表示敌人，显然用两个子弹就可以解决所有敌人。

解题思路

将输入的敌人的坐标
行坐标连向列坐标
如下图

求出最大匹配数
就可以得到最小点覆盖

代码

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
struct hhx{int to,next;
}a[10020];
int n,m,x,y,t,ans,p[520],b[520],head[520];
void add(int x,int y)
{a[++t].to=y;a[t].next=head[x];head[x]=t;
}
bool dfs(int d)
{for (int i=head[d];i;i=a[i].next)  //枚举匹配if (p[a[i].to]==0)  //没做过{p[a[i].to]=1;  //标记if (b[a[i].to]==0||dfs(b[a[i].to]))  //没匹配过，或者可以让出来{b[a[i].to]=d;  //匹配return 1;}}return 0;
}
int main()
{scanf("%d%d",&n,&m);for (int i=1;i<=m;i++){scanf("%d%d",&x,&y);add(x,y);  //连接}for (int i=1;i<=n;i++){memset(p,0,sizeof(p));  //清0if (dfs(i))  //求匹配ans++;}cout<<ans<<endl;return 0;
}