hdu 1151 Air Raid

A B C D E F G H I

D - Air Raid

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status

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

2433 41 32 3331 31 22 3

Sample Output

2
1
二分图：
最小点覆盖数=最大匹配数
增广路：
  若P是图G中一条连通两个未匹配顶点的路径，并且属于M的边和不属于M的边(即已匹配和待匹配的边)在P上交替出现，则称P为相对于M的一条增广路径。
 
　由增广路的定义可以推出下述三个结论：
   1、P的路径个数必定为奇数，第一条边和最后一条边都不属于M。
   2、将M和P进行取反操作可以得到一个更大的匹配M’
   （反操作：把P中的 匹配边 与 非匹配边 互换）
   3、M为G的最大匹配当且仅当不存在M的增广路径P
本题：
最小路径覆盖=图的顶点数-最大匹配数
二分图：
最小点覆盖数=最大匹配数
增广路：
  若P是图G中一条连通两个未匹配顶点的路径，并且属于M的边和不属于M的边(即已匹配和待匹配的边)在P上交替出现，则称P为相对于M的一条增广路径。
 
　由增广路的定义可以推出下述三个结论：
   1、P的路径个数必定为奇数，第一条边和最后一条边都不属于M。
   2、将M和P进行取反操作可以得到一个更大的匹配M’
   （反操作：把P中的 匹配边 与 非匹配边 互换）
   3、M为G的最大匹配当且仅当不存在M的增广路径P
本题：
最小路径覆盖=图的顶点数-最大匹配数
题目大意：
在一个城镇，有m个路口，和n条路，这些路都是单向的，而且路不会形成环，现在要弄一些伞兵去巡查这个城镇，
ACcode:
#include <cstring>#include <cstdio>#include <algorithm>#include <iostream>using namespace std;bool map[505][505];bool vis[505];int link[505];int m,n,gil,boy;bool DFS(int x){    for(int y=1; y<=n; ++y)    {        if(!vis[y] && map[x][y])        {            vis[y]=1;            if(!link[y] || DFS(link[y]))            {                link[y]=x;                return 1;            }        }    }    return 0;}int main(){    int i,x,y,T;    cin>>T;    while(T--)    {        scanf("%d%d",&n,&m);        memset(map,0,sizeof(map));        memset(link,0,sizeof(link));        int sum=0;        for(i=0; i<m; ++i)        {            scanf("%d%d",&x,&y);            map[x][y]=1;        }        for(x=1; x<=n; ++x)        {            memset(vis,0,sizeof(vis));            if(DFS(x))                ++sum;        }        printf("%d\n",n-sum);    }    return 0;}

二分图：

最小点覆盖数=最大匹配数

增广路：

  若P是图G中一条连通两个未匹配顶点的路径，并且属于M的边和不属于M的边(即已匹配和待匹配的边)在P上交替出现，则称P为相对于M的一条增广路径。

 

　由增广路的定义可以推出下述三个结论：

   1、P的路径个数必定为奇数，第一条边和最后一条边都不属于M。

   2、将M和P进行取反操作可以得到一个更大的匹配M’

   （反操作：把P中的 匹配边 与 非匹配边 互换）

   3、M为G的最大匹配当且仅当不存在M的增广路径P

本题：

最小路径覆盖=图的顶点数-最大匹配数