【POJ3041】Asteroids(二分图)
来源:互联网 发布:广通软件怎么样 编辑:程序博客网 时间:2024/06/02 04:19
http://poj.org/problem?id=3041
Asteroids
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 23472 Accepted: 12739
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.
OutputLine 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
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).
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,k;int vis[505];int fr[505][505];int f[505];bool find(int x){ for(int i=1;i<=n;i++) { if(!vis[i]&&fr[x][i]) { vis[i]=true; if(f[i]==-1) { f[i]=x; return true; } else if(find(f[i])) { f[i]=x; return true; } } } return false;}int main(){ while(~scanf("%d%d",&n,&k)) { memset(f,-1,sizeof(f)); memset(fr,false,sizeof(fr)); for(int i=1;i<=k;i++) { int a,b; scanf("%d%d",&a,&b); fr[a][b]=true; } int ans=0; for(int i=1;i<=n;i++) { memset(vis,false,sizeof(vis)); if(find(i)) ans++; } printf("%d\n",ans); }return 0;}
- 【POJ3041】Asteroids(二分图)
- POJ3041 Asteroids(二分图)
- Poj3041-Asteroids-【二分图】
- POJ3041-Asteroids(二分图匹配)
- POJ3041:Asteroids(二分图匹配)
- poj3041--Asteroids(二分匹配)
- POJ3041——Asteroids(二分图最大匹配)
- poj3041——Asteroids(二分图,匈牙利算法)
- poj3041 - Asteroids (二分图最小顶点覆盖)
- POJ3041--Asteroids--二分图最大匹配--Konig
- poj3041 Asteroids 最小点覆盖 二分图
- (二分图最大匹配) poj3041 Asteroids
- POJ3041 Asteroids【二分图最小点覆盖】
- Asteroids POJ3041 二分图最小顶点覆盖
- POJ3041--Asteroids(二分图,最小覆盖点)
- POJ3041 Asteroids【二分匹配】
- poj3041 Asteroids 二分图最小点集覆盖
- poj3041 Asteroids 最小点覆盖 二分图匹配
- LintCode:H-求K数和
- 柱状图js封装插件
- Hadoop+Spark学习
- 算法的复杂度
- 设计模式-工厂方法模式
- 【POJ3041】Asteroids(二分图)
- 2017多校训练Contest5: 1001 Rikka with Candies hdu6085
- Vue键盘事件
- 动态规划专题之石子合并
- [NOIP2012]同余方程
- null强制转任何类型的对象
- Android学习心得(三)关于真机调试后,签名的apk无法安装的问题
- WPF入门基础
- 基于commons-fileupload-1.2.jar的纯jsp文件上传