Air Raid
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 7511   Accepted: 4471

Description

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

Source

题解:伞兵可以访问所有路口的最少伞兵数量

Each paratrooper lands at an intersection and can visit other intersections following the town streets

只能访问两个路口

代码:

 #include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=;
int mp[MAXN][MAXN],vis[MAXN],usd[MAXN];
int N;
bool dfs(int u){
for(int i=;i<=N;i++){
if(!vis[i]&&mp[u][i]){
vis[i]=;
if(!usd[i]||dfs(usd[i])){
usd[i]=u;return true;
}
}
}
return false;
}
/*void floyd(){
for(int i=1;i<=N;i++)
for(int j=1;j<=N;j++)
if(mp[i][j]==0){
for(int k=1;k<=N;k++)
if(mp[i][k]&&mp[k][j])
mp[i][j]=1;
}
}*/
int main(){
int M,T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&N,&M);
int a,b;
mem(mp,);mem(usd,);
while(M--){
scanf("%d%d",&a,&b);
mp[a][b]=;
}
int ans=;
// floyd();
for(int i=;i<=N;i++){
mem(vis,);
if(dfs(i))ans++;
}
printf("%d\n",N-ans);
}
return ;
}

Air Raid(最小路径覆盖)的更多相关文章

  1. 【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 ...

  2. (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 ...

  3. 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 ...

  4. 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( ...

  5. (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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. 关于js封装框架类库之选择器引擎(一)

    选择器模块之传统做法 var tag = function (tag){ return document.getElementsByTagName(tag); } var id = function ...

  2. Cookie管理

    1,判断来访者是否第一次 public class VistorTest extends HttpServlet { @Override protected void doGet(HttpServle ...

  3. ubuntu 16.04环境配置

    ubuntu 16:1.源cp /etc/apt/sources.list /etc/apt/sources.list.bkpvi /etc/apt/sources.list-+{    deb ht ...

  4. Java 网络编程(一) 网络基础知识

    链接地址:http://www.cnblogs.com/mengdd/archive/2013/03/09/2951826.html 网络基础知识 网络编程的目的:直接或间接地通过网络协议与其他计算机 ...

  5. bzoj 1047 : [HAOI2007]理想的正方形 单调队列dp

    题目链接 1047: [HAOI2007]理想的正方形 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 2369  Solved: 1266[Submi ...

  6. [LeetCode]题解(python):085-Maximal Rectangle

    题目来源: https://leetcode.com/problems/maximal-rectangle/ 题意分析: 给定一个二维的二进制矩阵,也就是只包括0 和 1的,找出只包括1的最大的矩阵的 ...

  7. HTML+CSS笔记 CSS中级 一些小技巧

    水平居中 行内元素的水平居中 </a></li> <li><a href="#">2</a></li> &l ...

  8. 有意思的C宏

    在Linux内核.嵌入式代码等传统的C代码里,会有一些难以识别的宏定义.我记得在eCos, UBoot, FFmpeg有一些比较BT的宏定义,很难读懂.对于C++程序员来说,最好将这种难读的宏定义转成 ...

  9. asp.net mvc重写RequestValidator

    /// <summary> /// <httpRuntime requestValidationType="xxx.CustomRequestValidator" ...

  10. 帝国cms7.0 列表模板调用本栏目缩略图

    [e:loop={"select classimg from phome_enewsclass where classid='$GLOBALS[navclassid]'",1,24 ...