poj 3041 Asteroids

Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17963 Accepted: 9786

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 41 11 32 23 2

Sample Output

2

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).

把每一列当成一个点，每一行当成一个点，若行节点和列节点之间有边，则表明该行列该列有一个障碍物。

主要是构图：将每一行当成一个点，构成集合1， 每一列也当成一个点，构成集合2；每一个障碍物的位置坐标将集合1与集合2中的点连接起来，也就是将每一个障碍物作为连接节点的边。这样可以轻易的得出本题是一个最小点覆盖的问题，假设1个行节点覆盖了5个列节点，即这个行节点与这5个列节点间有5条边（即五个障碍物），由于这5条边都被那个行节点覆盖，即表明这5个障碍物都在同一列上，于是可以一颗炸弹全部清除，而本题也就转化成求最小点覆盖数的问题。

又有一个定理是：最小点覆盖数 = 最大匹配数， 所以此题转化成求最大匹配数。

以下是匈牙利算法的代码：

        

#include<stdio.h>#include<string.h>#define  N 1010int map[N][N] ,mode[N],vis[N];int n,m;int find(int x)//寻找增广路！ 找到返回1，否则返回0！{int i,j;for(i=1;i<=n;i++) {if(map[x][i]&&!vis[i])//与x相连点，并且没有遍历到{vis[i]=1;//标记为遍历过 if(mode[i]==0||find(mode[i]))//i 点 没有和另一部分匹配或 和i配对的点 没有匹配！ {mode[i]=x;return 1;}}}return 0;}int main(){int i,j,k;while(scanf("%d%d",&n,&m)!=EOF){memset(map,0,sizeof(map));memset(mode ,0,sizeof(mode));  for(i=0;i<m;i++)  {  int x,y;  scanf("%d%d",&x,&y);  if(x&&y)  map[x][y]=1;//标记为通路    }    int s=0;   for(i=1;i<=n;i++)    {    memset(vis,0,sizeof(vis));    if(find(i))//增广路     s++;}printf("%d\n",s); }  return 0;}