Description

Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting from an intersection and walking through town's streets you can never reach the same intersection
i.e. the town's streets form no cycles.



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

Your program should read sets of data. The first line of the input file contains the number of the data sets. Each data set specifies the structure of a town and has the format:




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

The result of the program is on standard output. For each input data set the program prints on a single line, starting from the beginning of the line, one integer: the minimum number of paratroopers required to visit all the intersections
in the town.

Sample Input

2
4
3
3 4
1 3
2 3
3
3
1 3
1 2
2 3

Sample Output

2
1

关于最小覆盖:http://blog.csdn.net/u014665013/article/details/49870029

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
using namespace std;
#define MAX 125
int n;
int map[MAX][MAX];
bool visited[MAX];
int match[MAX]; bool find(int i) ///查找当前的i是否可以匹配
{
int j;
for(j=1;j<=n;j++)
{
if(map[i][j]&&!visited[j])
{
visited[j]=1;
if(match[j]==-1||find(match[j]))
{
match[j]=i;
return 1;
}
}
}
return 0;
}
int main()
{
int k,x,y,ans;
int T;
scanf("%d",&T);
while(T--)
{
ans=0;
scanf("%d%d",&n,&k);
memset(map,0,sizeof(map));
memset(match,-1,sizeof(match)); for(int i=0;i<k;i++)//对有意思的进行初始化
{
scanf("%d%d",&x,&y);
map[x][y]=1;
}
for(int i=1;i<=n;i++)
{
memset(visited,0,sizeof(visited));//开始标记为全部没有访问
if(find(i))
ans++;
}
printf("%d\n",n-ans);
}
return 0;
}

Air Raid(最小路径覆盖)的更多相关文章

  1. 【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 ...

  2. (hdu step 6.3.3)Air Raid(最小路径覆盖:求用最少边把全部的顶点都覆盖)

    题目: Air Raid Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...

  3. HDU1151 Air Raid —— 最小路径覆盖

    题目链接:https://vjudge.net/problem/HDU-1151 Air Raid Time Limit: 2000/1000 MS (Java/Others)    Memory L ...

  4. POJ 1422 Air Raid (最小路径覆盖)

    题意 给定一个有向图,在这个图上的某些点上放伞兵,可以使伞兵可以走到图上所有的点.且每个点只被一个伞兵走一次.问至少放多少伞兵. 思路 裸的最小路径覆盖. °最小路径覆盖 [路径覆盖]在一个有向图G( ...

  5. (step6.3.4)hdu 1151(Air Raid——最小路径覆盖)

    题意:     一个镇里所有的路都是单向路且不会组成回路. 派一些伞兵去那个镇里,要到达所有的路口,有一些或者没有伞兵可以不去那些路口,只要其他人能完成这个任务.每个在一个路口着陆了的伞兵可以沿着街去 ...

  6. hdu 1151 Air Raid 最小路径覆盖

    题意:一个城镇有n个路口,m条路.每条路单向,且路无环.现在派遣伞兵去巡逻所有路口,伞兵只能沿着路走,且每个伞兵经过的路口不重合.求最少派遣的伞兵数量. 建图之后的就转化成邮箱无环图的最小路径覆盖问题 ...

  7. Air Raid(最小路径覆盖)

    Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7511   Accepted: 4471 Descript ...

  8. poj 1422 Air Raid 最少路径覆盖

    题目链接:http://poj.org/problem?id=1422 Consider a town where all the streets are one-way and each stree ...

  9. 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 ...

  10. POJ1422 Air Raid 【DAG最小路径覆盖】

    Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6763   Accepted: 4034 Descript ...

随机推荐

  1. SwitchyOmega

    SwitchyOmega下载安装地址: http://switchyomega.com/download.html GFWList.bak.txt教程 {"+GFWed":{&qu ...

  2. C#连接SQL Server数据库进行简单操作

    环境:VS2010 + SqlServer 2008 首先,按照面向对象的程序设计思想,设计一个数据库操作工具类MyTool.cs,该类中封装了关于数据库连接和操作的方法,各个功能模块在需进行数据库操 ...

  3. HackerRank "Morgan and a String"

    I saw the same sub-problem in LeetCode, and there exists a O(n) neat greedy solution: for _ in range ...

  4. bzoj4716 假摔

    Description [题目背景] 小Q最近喜欢上了一款游戏,名为<舰队connection>,在游戏中,小Q指挥强大的舰队南征北战,从而成为了一名 dalao.在游戏关卡的攻略中,可能 ...

  5. Centos7和win7双系统调整默认启动

    centos7之后都上grub2了,所以你要更改默认启动项什么的就不能像以前一样去改 /etc/grub.conf 当然你更不能去改/etc/grub2.conf 上了grub2之后,在设计有意规避让 ...

  6. SVN:通过Client端打tag

    教你如何使用svnClient打tag~给公司人用的! 1.进入代码主目录 2.右击空白处“TortoiseSVN”—->“Branch/tag” 3.点地址栏右侧的 (选择tags存放目录) ...

  7. (四)java程序基本组成

    一个基本的java程序一般包括几个部分,分别是程序所在的包名.程序中用到的其他包的路径.程序的类.类中的方法.变量和字面量. package demo; import java.util.Date; ...

  8. 黄聪:C#设置窗体打开位置(在显示器的右下角打开)

    ; ; this.SetDesktopLocation(x, y); 注释:System.Windows.Forms.Screen.PrimaryScreen.WorkingArea.Size.Wid ...

  9. SQL语言的三个分类:DDL、DML、DCL

    DML:数据操纵语言,主要是完成数据的新增,修改,删除和查询的操作. DDL:数据定义语言,主要是用来创建或修改表.视图.存储过程以及用户等. DCL:数据控制语言,是用来设置或更改数据库用户或角色权 ...

  10. Hbase与hive整合

    //hive与hbase整合create table lectrure.hbase_lecture10(sname string, score int) stored by 'org.apache.h ...