POJ1422 Air Raid
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 8006 | Accepted: 4803 |
Description
With these assumptions your task is to write a program that finds the minimum number of paratroopers that can descend on the town and visit all the intersections of this town in such a way that more than one paratrooper visits no intersection. Each paratrooper lands at an intersection and can visit other intersections following the town streets. There are no restrictions about the starting intersection for each paratrooper.
Input
no_of_intersections
no_of_streets
S1 E1
S2 E2
......
Sno_of_streets Eno_of_streets
The first line of each data set contains a positive integer no_of_intersections (greater than 0 and less or equal to 120), which is the number of intersections in the town. The second line contains a positive integer no_of_streets, which is the number of streets in the town. The next no_of_streets lines, one for each street in the town, are randomly ordered and represent the town's streets. The line corresponding to street k (k <= no_of_streets) consists of two positive integers, separated by one blank: Sk (1 <= Sk <= no_of_intersections) - the number of the intersection that is the start of the street, and Ek (1 <= Ek <= no_of_intersections) - the number of the intersection that is the end of the street. Intersections are represented by integers from 1 to no_of_intersections.
There are no blank lines between consecutive sets of data. Input data are correct.
Output
Sample Input
2
4
3
3 4
1 3
2 3
3
3
1 3
1 2
2 3
Sample Output
2
1
Source
最小路径覆盖。匈牙利算法。
/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mxn=;
int read(){
int x=,f=;char ch=getchar();
while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int T,n,m;
int mp[mxn][mxn];
int link[mxn];
bool vis[mxn];
bool find(int u){
for(int i=;i<=n;i++){
if(mp[u][i] && !vis[i]){
vis[i]=;
if(link[i]==- || find(link[i])){
link[i]=u;
return ;
}
}
}
return ;
}
int solve(){
int res=;
for(int i=;i<=n;i++){
memset(vis,,sizeof vis);
if(find(i))res++;
}
return res;
}
int main(){
T=read();
int i,j,u,v;
while(T--){
memset(mp,,sizeof mp);
memset(link,-,sizeof link);
n=read();m=read();
for(i=;i<=m;i++){
u=read();v=read();
mp[u][v]=;
}
int res=solve();
printf("%d\n",n-res);
}
return ;
}
POJ1422 Air Raid的更多相关文章
- POJ1422 Air Raid 【DAG最小路径覆盖】
Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6763 Accepted: 4034 Descript ...
- POJ1422 Air Raid 和 CH6902 Vani和Cl2捉迷藏
Air Raid Language:Default Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9547 A ...
- Air Raid[HDU1151]
Air RaidTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submis ...
- hdu1151 二分图(无回路有向图)的最小路径覆盖 Air Raid
欢迎参加——BestCoder周年纪念赛(高质量题目+多重奖励) Air Raid Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65 ...
- 【网络流24题----03】Air Raid最小路径覆盖
Air Raid Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Su ...
- hdu-----(1151)Air Raid(最小覆盖路径)
Air Raid Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Su ...
- hdu 1151 Air Raid(二分图最小路径覆盖)
http://acm.hdu.edu.cn/showproblem.php?pid=1151 Air Raid Time Limit: 1000MS Memory Limit: 10000K To ...
- HDOJ 1151 Air Raid
最小点覆盖 Air Raid Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) T ...
- Air Raid(最小路径覆盖)
Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7511 Accepted: 4471 Descript ...
随机推荐
- iOS 集成银联支付(绕过文档的坑,快速集成)-转
本文是投稿文章,作者:南栀倾寒当初集成支付宝的时候,觉得见了这么丑的代码,加上这么难找的下载地址,在配上几乎为零的文档,寒哥就要吐血了. 下午去集成银联,才知道血吐的早了. 下载地址:https:// ...
- 一个简单的公式——求小于N且与N互质的数的和
首先看一个简单的东西. 若$gcd(i,n)=1$,则有$gcd(n-i,n)=1$ 于是在小于$n$且与$n$互质的数中,$i$与$n-i$总是成对存在,且相加等于$n$. 考虑$i=n-i$的特殊 ...
- css的过渡背景色
css3新增的渐变背景色属性用法 原博客地址:http://caibaojian.com/css3-background-gradient.html
- 白话容器namespace
进入正题之前是例行装X环节: 过年7天吃的,花了45天又回来了. 近年来容器大火,也正是因为容器,生生灭掉了一个IT岗位!哥也是被生生的逼上了邪路. 那究竟什么是容器呢? 就三个字:它就是个进程!(多 ...
- Python学习 Day 1-简介 安装 Hello world
简介 Python(英语发音:/ˈpaɪθən/), 是一种面向对象.解释型计算机程序设计语言,由Guido van Rossum于1989年底发明,第一个公开发行版发行于1991年,Python 源 ...
- CentOS安装使用vnc进行远程桌面登录
以下介绍在CentOS 7下安装vncserver并使用vnc-viewer进行登录(使用root权限): 1.运行命令yum install tigervnc-server安装vncserver: ...
- 在Eclipse中用Maven打包jar包--完整版
将本地的jar导入到maven本地库中 <!--手动加入库中 --><!-- mvn install:install-file -DgroupId=org.apache.Hadoop ...
- 十一,类型参数化--Scala
类型参数化 在scala中,类型参数化(类似于泛型)使用方括号实现,如:Foo[A],同时,我们称Foo为高阶类型.如果一个高阶类型有2个类型参数,则在声明变量类型时可以使用中缀形式来表达,此时也称该 ...
- CSS hover 改变另外一个元素状态
Part.1 问题 我们写页面时也不少遇到这个问题,在没有使用任何预处理语言前提下,当hover 一个元素的时候怎么改变其它的元素? 这里我把它分为两种情况(除自身以外) hover时 1: 改变本身 ...
- DEBUG无法进入断点解决方法
18/08/17 任务栏:Tools->Options->Debugging->General->Require source files to exactly match t ...