*HDU1151 二分图
Air Raid
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4959 Accepted Submission(s): 3339
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.
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.
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.
1
// 模板 有向图最小路径覆盖=n-最大匹配。(用最小的路径覆盖所有的点)
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int mp[][],vis[],link[];
int n,Mu,Mv,m;
int dfs(int x)
{
for(int i=;i<=n;i++)
{
if(!vis[i]&&mp[x][i])
{
vis[i]=;
if(link[i]==-||dfs(link[i]))
{
link[i]=x;
return ;
}
}
}
return ;
}
int Maxcon()
{
int ans=;
memset(link,-,sizeof(link));
for(int i=;i<=n;i++)
{
memset(vis,,sizeof(vis));
if(dfs(i)) ans++;
}
return ans;
}
int main()
{
int t,a,b;
scanf("%d",&t);
while(t--)
{
memset(mp,,sizeof(mp));
scanf("%d%d",&n,&m);
for(int i=;i<m;i++)
{
scanf("%d%d",&a,&b);
mp[a][b]=;
}
Mu=Mv=n;
printf("%d\n",n-Maxcon());
}
return ;
}
*HDU1151 二分图的更多相关文章
- 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 ...
- hdu1151+poj2594(最小路径覆盖)
传送门:hdu1151 Air Raid 题意:在一个城镇,有m个路口,和n条路,这些路都是单向的,而且路不会形成环,现在要弄一些伞兵去巡查这个城镇,伞兵只能沿着路的方向走,问最少需要多少伞兵才能把所 ...
- C - NP-Hard Problem(二分图判定-染色法)
C - NP-Hard Problem Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:262144 ...
- POJ 2125 Destroying the Graph 二分图最小点权覆盖
Destroying The Graph Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8198 Accepted: 2 ...
- bzoj4025 二分图
支持加边和删边的二分图判定,分治并查集水之(表示我的LCT还很不熟--仅仅停留在极其简单的模板水平). 由于是带权并查集,并且不能路径压缩,所以对权值(到父亲距离的奇偶性)的维护要注意一下. 有一个小 ...
- hdu 1281 二分图最大匹配
对N个可以放棋子的点(X1,Y1),(x2,Y2)......(Xn,Yn);我们把它竖着排看看~(当然X1可以对多个点~) X1 Y1 X2 Y2 X3 Y3 ..... Xn Yn ...
- POJ 2226二分图最大匹配
匈牙利算法是由匈牙利数学家Edmonds于1965年提出,因而得名.匈牙利算法是基于Hall定理中充分性证明的思想,它是二部图匹配最常见的算法,该算法的核心就是寻找增广路径,它是一种用增广路径求二分图 ...
- 二分图&网络流&最小割等问题的总结
二分图基础: 最大匹配:匈牙利算法 最小点覆盖=最大匹配 最小边覆盖=总节点数-最大匹配 最大独立集=点数-最大匹配 网络流: 技巧: 1.拆点为边,即一个点有限制,可将其转化为边 BZOJ1066, ...
- POJ2195 Going Home[费用流|二分图最大权匹配]
Going Home Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22088 Accepted: 11155 Desc ...
随机推荐
- Java与MySQL的连接
下载数据库驱动文件,解压并保存至任意位置 下载地址 新建Java项目,并将驱动文件添加到项目中 项目名右键-->构建路径-->配置构建路径-->添加外部Jar 在项目中新建类,编写代 ...
- Java Web 学习链接
解决JSP中文乱码问题:http://www.cnblogs.com/chengkai/articles/2171848.html 编程思想之多线程与多进程:http://blog.csdn.net/ ...
- #define #include #undef的其中一个用法(目的)
一.背景 最近在跟一段系统级的代码,和原来单纯的下位机代码相比,真的是刘姥姥进大观园--看花了眼.相较于 之前所常见的各种下位机代码,系统级代码常常会出现深层次结构体嵌套,结构体内的各种回调函数导致对 ...
- java生产者/消费者模式实现——一生产者一消费者(操作值)
胶多不粘话多不甜,直接上代码: 生产者类: /** * Created by 51304 on 2016/2/28. */ public class P { private String lock; ...
- vue2.0学习(一)
1.解决双花括号在初始化时的闪烁,两种方式,一种是<div v-text="name"></div>,将用v-text指令来显示,类似于angular的ng ...
- 逻辑思维面试题-java后端面试-遁地龙卷风
(-1)写在前面 最近参加了一次面试,对笔试题很感兴趣,就回来百度一下.通过对这些题目的思考让我想起了建模中的关联,感觉这些题如果没接触就是从0到1,考验逻辑思维的话从1到100会更好,并且编程简易模 ...
- 1.reset.css的设置
/* reset css */ *, ::before, ::after{ /*选择所有的标签 */ margin: 0; padding: 0; /*清除移动端默认的 点击高亮效果*/ -webki ...
- ServletConfig 可以做啥
1.获得 servlet配置的servletname 2.获得servlet 配置的 getInitParameter("keyname") 3.获得servlet配置的 所有的 ...
- 继承Thread类
Thread类在包java.lang中,从这个类中实例化的对象代表线程,启动一个新线程需要建立Thread实例,Thread类中常用的两个构造方法如下: (1)public Thread(String ...
- 如何从零基础学习VR
转载请声明转载地址:http://www.cnblogs.com/Rodolfo/,违者必究. 近期很多搞技术的朋友问我,如何步入VR的圈子?如何从零基础系统性的学习VR技术? 本人将于2017年1月 ...