http://poj.org/problem?id=1422

Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 8579   Accepted: 5129

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

 
最短路径覆盖==总点数-最大匹配数
 #include <cstring>
#include <cstdio> using namespace std; int n,map[][],match[];
bool vis[]; bool find(int u)
{
for(int v=;v<=n;v++)
if(map[u][v]&&!vis[v])
{
vis[v]=;
if(!match[v]||find(match[v]))
{
match[v]=u;
return true;
}
}
return false;
} int AC()
{
int t;scanf("%d",&t);
for(int m,ans;t--;)
{
scanf("%d%d",&n,&m);ans=n;
for(int u,v;m--;map[u][v]=)
scanf("%d%d",&u,&v);
for(int i=;i<=n;i++)
{
memset(vis,,sizeof(vis));
if(find(i)) ans--;
}
printf("%d\n",ans);
memset(map,,sizeof(map));
memset(match,,sizeof(match));
}
return ;
} int I_want_AC=AC();
int main(){;}

POJ——T 1422 Air Raid的更多相关文章

  1. 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 ...

  2. poj 1422 Air Raid (二分匹配)

    Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6520   Accepted: 3877 Descript ...

  3. poj——1422 Air Raid

    Air Raid Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8577   Accepted: 5127 Descript ...

  4. 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 ...

  5. POJ 1422 Air Raid

    题目链接: http://poj.org/problem?id=1422 Description Consider a town where all the streets are one-way a ...

  6. 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( ...

  7. POJ - 1422 Air Raid 二分图最大匹配

    题目大意:有n个点,m条单向线段.如今问要从几个点出发才干遍历到全部的点 解题思路:二分图最大匹配,仅仅要一条匹配,就表示两个点联通,两个点联通仅仅须要选取当中一个点就可以,所以有多少条匹配.就能够减 ...

  8. POJ - 1422 Air Raid(DAG的最小路径覆盖数)

    1.一个有向无环图(DAG),M个点,K条有向边,求DAG的最小路径覆盖数 2.DAG的最小路径覆盖数=DAG图中的节点数-相应二分图中的最大匹配数 3. /* 顶点编号从0开始的 邻接矩阵(匈牙利算 ...

  9. 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 ...

随机推荐

  1. HDU 1086 You can Solve a Geometry Problem too( 判断线段是否相交 水题 )

    链接:传送门 题意:给出 n 个线段找到交点个数 思路:数据量小,直接暴力判断所有线段是否相交 /*************************************************** ...

  2. Myeclipse关闭JS等文件的验证

    点击 window > 右键单击properties,弹出properties界面 然后选择MyEclipse->validation->Excluded Resource下找到不需 ...

  3. IE9以下版本兼容h5标签

    随着html5(后面用h5代表)标签越来越广泛的使用,IE9以下(IE6-IE8)不识别h5标签的问题让人很是烦恼. 在火狐和chrome之类的浏览器中,遇到不认识的标签,只要给个display:bl ...

  4. Java基础学习总结(48)——Java 文档注释

    Java只是三种注释方式.前两种分别是// 和/* */,第三种被称作说明注释,它以/** 开始,以 */结束. 说明注释允许你在程序中嵌入关于程序的信息.你可以使用javadoc工具软件来生成信息, ...

  5. 洛谷——P1886 滑动窗口|| POJ——T2823 Sliding Window

    https://www.luogu.org/problem/show?pid=1886#sub || http://poj.org/problem?id=2823 题目描述 现在有一堆数字共N个数字( ...

  6. HDU 4318 Contest 2

    简单的一题,使用类DIJK的算法就可以了. #include <iostream> #include <cstdio> #include <queue> #incl ...

  7. [Hyperapp] Interact with the State Object through Hyperapp Action functions

    Hyperapp is an ultra lightweight (1kb), minimal, functional, JavaScript library for building UIs. It ...

  8. 51nod-1346: 递归

    [传送门:51nod-1346] 简要题意: 给出一个式子a[i][j]=a[i-1][j]^a[i][j-1] 给出a[1][i],a[i][1](2<=i<=131172) 有n个询问 ...

  9. Kettle学习系列之kettle的下载、安装和初步使用(windows平台下)(图文详解)

    不多说,直接上干货! kettle的下载 žKettle可以在http://kettle.pentaho.org/网站下载                   http://sourceforge.n ...

  10. 固定执行计划-使用coe_xfr_sql_profile

    一.历史执行计划固定 历史的执行计划找到一个合理的执行计划进行绑定 1. 存在多个执行计划的语句,按照索引是比较合适的,FULL SCAN不合适 select * from scott.emp whe ...