P - Air Raid
来源poj1422
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
要多少个人才能走完所有路,最小路径覆盖,用匈牙利算法,最小路=总节点-最大匹配
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+100;
const double eps=1e-8;
using namespace std;
const double pi=acos(-1.0);
const int inf=0xfffffff;
const int N=220;
int pre[N];
int visit[N],line[N][N];
char Map[N][N];
int n,m,y,x;
bool find(int x)
{
rep(i,1,n+1)
{
if(line[x][i]&&visit[i]==0)
{
visit[i]=1;
if(pre[i]==0||find(pre[i]))
{
pre[i]=x;
return true;
}
}
}
return false;
}
int main()
{
int re;
cin>>re;
while(re--)
{
mm(line,0);
mm(pre,0);
cin>>n>>m;
rep(i,0,m)
{
cin>>x>>y;
line[x][y]=1;
}
int ans=0;
rep(i,1,n+1)
{
mm(visit,0);
if(find(i)) ans++;
}
cout<<n-ans<<endl;
}
return 0;
}
P - Air Raid的更多相关文章
- Air Raid[HDU1151]
Air RaidTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submis ...
- 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 ...
- 【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-----(1151)Air Raid(最小覆盖路径)
Air Raid Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Su ...
- 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 ...
- 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) T ...
- Air Raid(最小路径覆盖)
Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7511 Accepted: 4471 Descript ...
- POJ1422 Air Raid 【DAG最小路径覆盖】
Air Raid Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6763 Accepted: 4034 Descript ...
- POJ 1422 Air Raid(二分图匹配最小路径覆盖)
POJ 1422 Air Raid 题目链接 题意:给定一个有向图,在这个图上的某些点上放伞兵,能够使伞兵能够走到图上全部的点.且每一个点仅仅被一个伞兵走一次.问至少放多少伞兵 思路:二分图的最小路径 ...
- Air Raid
Air Raid Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Subm ...
随机推荐
- Metadata获取的三种方式
本文的试验环境为CentOS 7.3,Kubernetes集群为1.11.2,安装步骤参见kubeadm安装kubernetes V1.11.1 集群 0. Metadata 每个Pod都有一些信息, ...
- docker dcm4chee
The received images should show up in the UI of the Archive at http://localhost:8080/dcm4chee-arc/ui ...
- 网络编程之 keepalive(zz)
link1: http://tldp.org/HOWTO/html_single/TCP-Keepalive-HOWTO/ link2: http://dev.csdn.net/article/849 ...
- Spark2.3(四十):如何使用java通过yarn api调度spark app,并根据appId监控任务,关闭任务,获取任务日志
背景: 调研过OOZIE和AZKABA,这种都是只是使用spark-submit.sh来提交任务,任务提交上去之后获取不到ApplicationId,更无法跟踪spark application的任务 ...
- BABLE 原理
1.babel转换原理 2.主要过程 (1)babylon进行解析得到AST (2)babel-traverse插件对AST树进行遍历转译得到新的AST树 (3)babel-generator将AST ...
- 【PHP】解析PHP中的错误和异常处理
目录结构: contents structure [-] 错误级别 自定义处理器 设置异常日志 自定义异常类 在这篇文章中,笔者将会阐述PHP中的异常处理,希望能够对你有所帮助. 1.错误级别 PHP ...
- CentOS 6.5 x64相关安全,优化配置
一.安全 1.修改密码长度: [root@CentOS64 ~]# vi /etc/login.defs PASS_MAX_DAYS 99999 //用户的密密码最长使用天数 PASS_MIN_D ...
- [转]Redis cluster failover
今天测试了redis cluster failover 功能,在切换过程中很快,但在failover时有force 与takeover 之分 [RHZYTEST_10:REDIS:6237:M ~] ...
- 戳破ZigBee技术智能家居的谎言!
戳破ZigBee技术智能家居的谎言 一.ZigBee介绍 简介 在蓝牙技术的使用过程中,人们发现蓝牙技术尽管有许多优点,但仍存在许多缺陷.对工业,家庭自动化控制和遥测遥控领域而言,蓝牙技术显得太复杂, ...
- Android中的指纹识别
转载请注明出处:http://blog.csdn.net/wl9739/article/details/52444671 评论中非常多朋友反映,依据我给出的方案,拿不到指纹信息这个问题,在这里统一说明 ...