Air Raid
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 题意:有几个伞兵降落,需要走完所有的路去侦查,每条路只能允许一个伞兵经过一次,也就是不想交
然后问你最少需要几个伞兵。
 #include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
const int N=; int cas,n,m;
int cnt,head[N],next[N],rea[N];
int flag[N],ly[N]; void add(int u,int v){next[++cnt]=head[u],head[u]=cnt,rea[cnt]=v;}
bool dfs(int u)
{
for (int i=head[u];i!=-;i=next[i])
{
int v=rea[i];
if (flag[v]) continue;
flag[v]=;
if (!ly[v]||dfs(ly[v]))
{
ly[v]=u;
return ;
}
}
return ;
}
int main()
{
scanf("%d",&cas);
while (cas--)
{
scanf("%d%d",&n,&m);
cnt=;
memset(head,-,sizeof(head));
int x,y;
for (int i=;i<=m;i++)
{
scanf("%d%d",&x,&y);
add(x,y);
}
int ans=;
memset(ly,,sizeof(ly));
for (int i=;i<=n;i++)
{
memset(flag,,sizeof(flag));
ans+=dfs(i);
}
ans=n-ans;
printf("%d\n",ans);
}
}

打了10分钟的代码。

 

POJ1422Air Raid(二分图,最小不相交路径覆盖)的更多相关文章

  1. P2172 [国家集训队]部落战争 二分图最小不相交路径覆盖

    二分图最小不相交路径覆盖 #include<bits/stdc++.h> using namespace std; ; ; ; ], nxt[MAXM << ], f[MAXM ...

  2. Air Raid POJ - 1422 【有向无环图(DAG)的最小路径覆盖【最小不相交路径覆盖】 模板题】

    Consider a town where all the streets are one-way and each street leads from one intersection to ano ...

  3. [luoguP2765] 魔术球问题(最大流—最小不相交路径覆盖)

    传送门 枚举球的个数 num 如果 i < j && (i + j) 是完全平方数,那么 i -> j' 连一条边 再加一个超级源点 s,s -> i 再加一个超级汇 ...

  4. POJ Air Raid 【DAG的最小不相交路径覆盖】

    传送门:http://poj.org/problem?id=1422 Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissi ...

  5. HDU 4862 Jump(最小K路径覆盖)

    输入一个n×m网格图,每个结点的值为0-9,可以从任意点出发不超过k次,走完每个点且仅访问每个结点一次,问最终的能量最大值.不可全部走完的情况输出-1. 初始能量为0. 而结点(x,y)可以跳跃到结点 ...

  6. 网络费用流-最小k路径覆盖

    多校联赛第一场(hdu4862) Jump Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Ot ...

  7. 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 ...

  8. POJ2125 Destroying The Graph(二分图最小点权覆盖集)

    最小点权覆盖就是,对于有点权的有向图,选出权值和最少的点的集合覆盖所有的边. 解二分图最小点权覆盖集可以用最小割: vs-X-Y-vt这样连边,vs和X部点的连边容量为X部点的权值,Y部和vt连边容量 ...

  9. POJ 2125 Destroying The Graph (二分图最小点权覆盖集+输出最小割方案)

    题意 有一个图, 两种操作,一种是删除某点的所有出边,一种是删除某点的所有入边,各个点的不同操作分别有一个花费,现在我们想把这个图的边都删除掉,需要的最小花费是多少. 思路 很明显的二分图最小点权覆盖 ...

随机推荐

  1. jmeter日记

    1.不要像loadrunner一样使用集合点,jmeter不需要 2.性能测试 线程组设置的时候 永远 3.启动jmeter之后,打开log 4.分布式的话 tps是多台电脑的tps之和 自动加好了的 ...

  2. django模板(过滤器)

    -------------------django内建的过滤器-------------------1.add 使用形式为:{{ value | add: "2"}} 意义:将va ...

  3. 很全面的Android面试题

    Activity 什么是Activity 四大组件之一,一个和用户交的互界面就是一个activity,是所有 View 的容器 我开发常用的的有FragmentActivitiy,ListActivi ...

  4. 三种计算c#程序运行时间的方法

    三种计算c#程序运行时间的方法 第一种: 利用 System.DateTime.Now // example1: System.DateTime.Now method DateTime dt1 = S ...

  5. 201421123059 http://www.cnblogs.com/deng201421123059/

    201421123059 http://www.cnblogs.com/deng201421123059/

  6. 扫雷游戏制作过程(C#描述):第五节、菜单操作(续)

    前言 这里给出教程原文地址. 该项目已经放在github上托管. 发布版已经分享到百度网盘 菜单操作(续) 接着节前一章节的内容,我们继续完善菜单栏的功能. 我们首先,先完善Rank的选项,我们希望我 ...

  7. Swing-setAlignmentX()用法-入门

    先看下API: public void setAlignmentX(float alignmentX) 设置垂直对齐方式. 参数: alignmentX - 新的垂直对齐方式 网上关于这个函数的详细情 ...

  8. 201521123117 《Java程序设计》第7周学习总结

    1. 本周学习总结 2.书面作业 Q1.ArrayList代码分析 1.解释ArrayList的contains源代码 源代码: //contains()方法 public boolean conta ...

  9. 201521123097《Java程序设计》第五周学习总结

    1. 本周学习总结 1.1 尝试使用思维导图总结有关多态与接口的知识点. 2. 书面作业 1.代码阅读:Child压缩包内源代码 1.1 com.parent包中Child.java文件能否编译通过? ...

  10. JAVA课程设计个人博客 学生成绩管理 201521123001 张陈东芳

    1. 团队课程设计博客链接 http://www.cnblogs.com/kawajiang/p/7062407.html 2.个人负责模块或任务说明 我主要负责实现学生信息的添加功能.学生成绩的录入 ...