http://www.lydsy.com/JudgeOnline/problem.php?id=1693

裸匈牙利。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; } const int N=505;
int ly[N+N], vis[N+N], ihead[N], cnt, n, m;
struct dat { int to, next; }e[10005];
void add(int u, int v) {
e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v;
}
bool ifind(int x) {
int y;
for(int i=ihead[x]; i; i=e[i].next) if(!vis[y=e[i].to]) {
vis[y]=1;
if(!ly[y] || ifind(ly[y])) {
ly[y]=x;
return true;
}
}
return false;
} int main() {
read(n); read(m);
for1(i, 1, m) {
int u=getint(), v=getint();
add(u, v+n);
}
int ans=0;
for1(i, 1, n) {
CC(vis, 0);
if(ifind(i)) ++ans;
}
print(ans);
return 0;
}

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

INPUT DETAILS:

The following diagram represents the data, where "X" is an
asteroid and "." is empty space:
X.X
.X.
.X.

Sample Output

2

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).

HINT

Source

【BZOJ】1693: [Usaco2007 Demo]Asteroids(匈牙利)的更多相关文章

  1. BZOJ1693: [Usaco2007 Demo]Asteroids

    n<=500 *n的格子,给m<=10000个格子有人,一炮可以清掉一行或一列的人(莫名的爽!)求最少几炮干掉所有人. 经典二分图模型!行成点,列成点,一个点就连接一行一列,表示这一行或这 ...

  2. BZOJ 1695 [Usaco2007 Demo]Walk the Talk 链表+数学

    题意:链接 方法:乱搞 解析: 出这道题的人存心报复社会. 首先这个单词表-先上网上找这个单词表- 反正总共2265个单词.然后就考虑怎么做即可了. 刚開始我没看表,找不到怎么做,最快的方法我也仅仅是 ...

  3. BZOJ 1629: [Usaco2007 Demo]Cow Acrobats

    Description Farmer John's N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away a ...

  4. BZOJ 1628 [Usaco2007 Demo]City skyline:单调栈

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1628 题意: 题解: 单调栈. 单调性: 栈内元素高度递增. 一旦出现比栈顶小的元素,则表 ...

  5. bzoj 1630: [Usaco2007 Demo]Ant Counting【dp】

    满脑子组合数学,根本没想到dp 设f[i][j]为前i只蚂蚁,选出j只的方案数,初始状态为f[0][0]=1 转移为 \[ f[i][j]=\sum_{k=0}^{a[i]}f[i-1][j-k] \ ...

  6. bzoj 1628: [Usaco2007 Demo]City skyline【贪心+单调栈】

    还以为是dp呢 首先默认答案是n 对于一个影子,如果前边的影子比它高则可以归进前面的影子,高处的一段单算: 和他一样高的话就不用单算了,ans--: 否则入栈 #include<iostream ...

  7. bzoj 1629: [Usaco2007 Demo]Cow Acrobats【贪心+排序】

    仿佛学到了贪心的新姿势-- 考虑相邻两头牛,交换它们对其他牛不产生影响,所以如果交换这两头牛能使这两头牛之间的最大值变小,则交换 #include<iostream> #include&l ...

  8. BZOJ1629: [Usaco2007 Demo]Cow Acrobats

    1629: [Usaco2007 Demo]Cow Acrobats Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 601  Solved: 305[Su ...

  9. BZOJ1628: [Usaco2007 Demo]City skyline

    1628: [Usaco2007 Demo]City skyline Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 256  Solved: 210[Su ...

随机推荐

  1. vagrant box各种命令汇总

    最近在研究laravel,中间用到了vagrant 虚拟机管理工具,学习一下他的命令 vagrant box命令 用于管理boxes的命令,比如添加.删除等等. 此命令的功能主要通过以下子命令完成: ...

  2. MySQL 连接方式

    MySQL 连接方式 1:TCP/IP 套接字方式 这种方式会在TCP/IP 连接上建立一个基于网络的连接请求,一般是client连接跑在Server上的MySQL实例,2台机器通过一个TCP/IP ...

  3. js 字符串indexof与search方法的区别

    1.indexof方法 indexOf() 方法可返回某个指定的字符串值在字符串中首次出现的位置. 语法: 注意:有可选的参数(即设置开始的检索位置). 2.search方法 search() 方法用 ...

  4. (数据库)DBCP连接池配置参数说明

    <!-- 数据源1 --> <bean id="dataSource" class="org.apache.commons.dbcp.BasicData ...

  5. ubuntu PATH 出错修复

    我的 ubuntu10.10设置交叉编译环境时,PATH 设置错误了,导致无法正常启动,错误情况如下: { PATH:找不到命令ubuntu2010@ubuntu:~$ ls命令 'ls' 可在 '/ ...

  6. STS(Spring Tool Suite)设置支持maven

  7. 视频播放器控制原理:ffmpeg之ffplay播放器源代码分析

    版权声明:本文由张坤原创文章,转载请注明出处: 文章原文链接:https://www.qcloud.com/community/article/535574001486630869 来源:腾云阁 ht ...

  8. C#: 数字经纬度和度分秒经纬度间的转换

    using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Cons ...

  9. python 火车票爬取代码

    1.根据搜索词下载百度图片: # -*- coding: utf-8 -*- """根据搜索词下载百度图片""" import re imp ...

  10. Faster RCNN原理分析 :Region Proposal Networks详解

    博主的论文笔记: https://blog.csdn.net/YZXnuaa/article/details/79221189 很详细! 另外,关于博主的博客很多拓展知识面: 120篇 深度学习23篇 ...