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)
Total Submission(s): 3002 Accepted Submission(s): 1951
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.
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.
2
4
3
3 4
1 3
2 3
3
3
1 3
1 2
2 3
2
1
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; const int maxn=220; bool mp[maxn][maxn],used[maxn];
int linker[maxn],n,m; bool dfs(int u)
{
for(int i=1;i<=n;i++)
{
if(used[i]) continue;
if(mp[u][i])
{
used[i]=true;
if(linker[i]==-1||dfs(linker[i])==true)
{
linker[i]=u;
return true;
}
}
}
return false;
} int hungary()
{
int ret=0;
memset(linker,-1,sizeof(linker));
for(int i=1;i<=n;i++)
{
memset(used,false,sizeof(used));
if(dfs(i)) ret++;
}
return ret;
} int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&m);
memset(mp,false,sizeof(mp));
for(int i=0;i<m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
mp[a][b]=true;
}
printf("%d\n",n-hungary());
}
return 0;
}
HDOJ 1151 Air Raid的更多相关文章
- 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 ...
- hdu 1150 Machine Schedule hdu 1151 Air Raid 匈牙利模版
//两道大水……哦不 两道结论题 结论:二部图的最小覆盖数=二部图的最大匹配数 有向图的最小覆盖数=节点数-二部图的最大匹配数 //hdu 1150 #include<cstdio> #i ...
- hdu 1151 Air Raid DAG最小边覆盖 最大二分匹配
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1151 题目大意: 城镇之间互相有边,但都是单向的,并且不会构成环,现在派伞兵降落去遍历城镇,问最少最少 ...
- hdu 1151 Air Raid - 二分匹配
Consider a town where all the streets are one-way and each street leads from one intersection to ano ...
- hdu - 1151 Air Raid(有向无环图的最小路径覆盖)
http://acm.hdu.edu.cn/showproblem.php?pid=1151 在一个城市里有n个地点和k条道路,道路都是单向的,并且不存在环.(DAG) 现在伞兵需要去n个地点视察,伞 ...
- (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)
题目大意: 有n个城市,m条道路,城市的道路是单向. 现在我们的伞兵要降落在城市里,然后我门的伞兵要搜索所有道路.问我们最少占领多少个城市就可以搜索所有的道路了. 我们可以沿着道路向前走到达另一个城 ...
- HDU 1151 - Air Raid
很明显求最小路径覆盖 就是求最大匹配 #include <iostream> #include <cstdio> #include <cstring> #inclu ...
- 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 ...
随机推荐
- JavaScript在IE和Firefox(火狐)的不兼容问题解决方法小结 【转】http://blog.csdn.net/uniqer/article/details/7789104
1.兼容firefox的 outerHTML,FF中没有outerHtml的方法. 代码如下: if (window.HTMLElement) { HTMLElement.prototype.__de ...
- Oracle DBA 的常用Unix参考手册(二)
9.AIX下显示CPU数量 # lsdev -C|grep Process|wc -l10.Solaris下显示CPU数量# psrinfo -v|grep "Status of pr ...
- win下Java环境安装
1.eclipse:eclipse.org 解压后直接打开 2.JDK:http://www.oracle.com/technetwork/java/javase/downloads/jdk7-do ...
- wifi详解(四)
1 IOCTL的调用逻辑 之所以要分析这个,是因为上层wpa_supplicant和WIFI驱动打交道的方式,多半是通过ioctl的方式进行的,所以看看它的调用逻辑(这里只列出其主要的调 ...
- poj 1651 http://poj.org/problem?id=1651
http://poj.org/problem?id=1651Multiplication Puzzle Time Limit: 1000MS Memory Limit: 65536K To ...
- hdu1722 bjfu1258 辗转相除法
这题就是个公式,代码极简单.但我想,真正明白这题原理的人并不多.很多人只是随便网上一搜,找到公式a了就行,其实这样对自己几乎没有提高. 鉴于网上关于这题的解题报告中几乎没有讲解原理的,我就多说几句,也 ...
- 仿酷狗音乐播放器开发日志二十七 用ole为窗体增加文件拖动功能(附源码)
转载请说明原出处,谢谢~~ 中秋到了,出去玩了几天.今天把仿酷狗程序做了收尾,已经开发完成了,下一篇博客把完结的情况说一下.在这篇博客里说一下使用OLE为窗体增加文件拖拽的功能.使用播放器,我更喜欢直 ...
- 2.1……Android中的单位简介
引用自Google API Guides Dimension A dimension value defined in XML. A dimension is specified with a num ...
- bzoj 3218 a + b Problem(最小割+主席树)
[题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=3218 [题意] 给n个格子涂白或黑色,白则wi,黑则bi的好看度,若黑格i存在: 1& ...
- MFC特定函数的应用20160720(SystemParametersInfo,GetWindowRect,WriteProfileString,GetSystemMetrics)
1.SystemParametersInfo函数可以获取和设置数量众多的windows系统参数 MFC中可以用 SystemParametersInfo(……) 函数来获取和设置系统信息,如下面例子所 ...