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. eclipse 怎么新建工作空间workspace

    打开eclipse 点击文件“File”菜单切换工作空间“Switch Workspace”>其它“Other” 点击“Browser”选择新的工作空间目录. 选择新的工作空间目录,点击确定. ...

  2. Android Studio如何添加override

    而Android Studio如何添加呢?方法如下: 右键(或者Alt + Insert) ---  Generate... ---- Override Method...  或者 Implement ...

  3. 修改oracle数据库密码

    1.用Xshell远程连接安装数据库的服务器,切换到安装oracle数据库的用户下,(我的oracle数据库就安装在oracle用户下) 命令: su - oracle; 2.进入oracle控制台 ...

  4. nginx ssi 配置小细节(一)

    最近工作需要使用nginx的ssi (server side include)技术,在这里,将使用中的一点心得分享一下,也是一种备忘! 首先,nginx的ssi启用很简单,就只有三个最基本的指令: s ...

  5. c# 鼠标在控件上拖动 移动窗体 移动窗口

    #region 移动窗体 移动窗口 private Point _mousePoint; private int topA(Control cc) { if (cc == null || cc == ...

  6. Redis 宣言(Redis Manifesto)

    Redis 的作者 antirez(Salvatore Sanfilippo)曾经发表了一篇名为 Redis 宣言(Redis Manifesto)的文章,文中列举了 Redis 的七个原则,以向大家 ...

  7. php反射机制获取未知类的详细信息

    使用ReflectionClass就可以获取未知类的详细信息 demo: require("hello.php"); $class = new ReflectionClass(&q ...

  8. python 深入理解 赋值、引用、拷贝、作用域

    在 python 中赋值语句总是建立对象的引用值,而不是复制对象.因此,python 变量更像是指针,而不是数据存储区域, 这点和大多数 OO 语言类似吧,比如 C++.java 等 ~ 1.先来看个 ...

  9. Linux xargs命令

    xargs是给命令传递参数的一个过滤器,也是组合多个命令的一个工具.它把一个数据流分割为一些足够小的块,以方便过滤器和命令进行处理.通常情况下,xargs从管道或者stdin中读取数据,但是它也能够从 ...

  10. Report_客制化以PLSQL输出XLS标记实现Excel报表(案例)

    2015-02-12 Created By BaoXinjian