Air Raid(最小路径覆盖)
Description
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
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
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
- 【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 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 ...
- 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 ...
- 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( ...
- (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 ...
- 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 ...
- Air Raid(最小路径覆盖)
Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7511 Accepted: 4471 Descript ...
- 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 ...
- 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 ...
- 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
- JavaScript:改变li前缀图片和样式
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/stri ...
- serialVersionUID 的用途--转加自己的疑问
serialVersionUID适用于Java的序列化机制.简单来说,Java的序列化机制是通过判断类的serialVersionUID来验证版本一致性的.在进行反序列化时,JVM会把传来的字节流中的 ...
- Hibernate3回顾-6-hibernate缓存(性能优化策略)
主要来源: http://blog.csdn.net/csh624366188/article/details/7612142 (比较详细) http://www.cnblogs.com/20091 ...
- Servlet Filter 1
1.Filter简介 )Filter也称之为过滤器,它是Servlet技术中最实用的技术,WEB开发人员通过Filter技术,对web服务器管理的所有web资源:例如Jsp, Servlet, 静态图 ...
- 05 Linux下开发JSP项目(Hello world)
测试环境: 主机系统:Win 7 虚拟机:VMware workstation 11.1.0 虚拟机OS: centos 6.5 64位 Kernel 2.6.32-431-e16.x86_64 My ...
- Silverlight开源框架SL提供便捷的二次开发银光框架
Silverlight开发框架SilverFrame欢迎咨询 基于Silverlight4.0开发,兼容Silverlight 5.0,SQLServer2005数据库.WCF: 本框架有清爽的前端界 ...
- Python基础教程【读书笔记】 - 2016/7/10
希望通过博客园持续的更新,分享和记录Python基础知识到高级应用的点点滴滴! 第五波:第1章 基础知识 [总览] 介绍如何得到所需的软件,然后讲一点点算法及其主要的组成.学习变量variable ...
- bzoj3136
Description 给定m个素数和Q个询问.每个询问有n个人,每次操作可以任意选择其中的一个素数p(素数可以重复使用),然后去掉剩余人数 mod p个人.对于每个询问,我们想知道,至少需要多少步操 ...
- WeX5和BeX5比较
http://wex5.com/cn/wex5和bex5比较/ WeX5和BeX5比较 许多对WeX5和BeX5略有了解得人都知道,WeX5和BeX5是完全共用前端框架技术的.但是WeX5和BeX5是 ...
- windows初始化后做了哪些事情
保护视力,将WINDOWS背景色变成淡绿色 绿色和蓝色对眼睛最好,建议大家在长时间用电脑后,经常看看蓝天.绿地,就能在一定程度上缓解视疲劳.同样的道理,如果我们把电脑屏幕和网页的底色变为淡淡的苹果绿, ...