hdu1151(二分匹配最小路径覆盖)

http://acm.hdu.edu.cn/showproblem.php?pid=1151

Air Raid

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)

Total Submission(s): 1966 Accepted Submission(s):
1261


Problem 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
Asia 2002, Dhaka (Bengal)



Recommend
Ignatius.L
最小边覆盖=最大独立集=n-最大匹配

题意:

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，

伞兵只能沿着路的方向走，问最少需要多少伞兵才能把所有的路口搜一遍。

解题:

(其实就是给一个m个点n条边的有向无环图，求该图的最小路径覆盖) 最小路径覆盖数=顶点数-最大匹配数。

下面解释一下:最小路径覆盖数=顶点数-最大匹配数。一条路径只能覆盖两个点。最大匹配多少，你就想像成

减去2*最大匹配数的点 ,剩下的每个点必须用一条边去覆盖最大匹配数*2,就是这最大匹配这么多条边,覆盖

掉的点 剩下的那些点，一个点必须要有一条边覆盖,

所以有： 最大匹配数+（n-2*最大匹配数 ）=最小路径覆盖





还是uN和vN这个地方老是出错啊。。wa了无数次，最后重新写了一遍再wa了一次然后ac了。。

#include

#include

#include

using namespace std;

const int maxn=1300;

int uN,vN,k;

int xM[maxn],yM[maxn];

bool chk[maxn];

int g[130][130];

//int n,m,k;

bool SearchPath(int u)

{

int v;


for(v=1;v<=uN;v++)//uN啊

{


if(g[u][v]&&!chk[v])


{


chk[v]=true;


if(yM[v]==-1||SearchPath(yM[v]))


{


yM[v]=u;


xM[u]=v;


return true;


}


}

}

return
false;

}

int MaxMatch()

{

int
u,ret=0;


memset(xM,-1,sizeof(xM));


memset(yM,-1,sizeof(yM));


for(u=1;u<=uN;u++)

{


if(xM[u]==-1)


{


memset(chk,false,sizeof(chk));


if(SearchPath(u))


{


ret++;


}


}

}

return
ret;

}

int main()

{

//int
n,k;

int t;


scanf("%d",&t);


while(t--)

{


//scanf("%d%d",&m,&k);


// memset(chk,false,sizeof(chk));


scanf("%d%d",&uN,&vN);


//uN=n;


//vN=m;


int x,y;


memset(g,0,sizeof(g));


// int i;


for(int j=0;j


{


scanf("%d%d",&x,&y);


//if(g[x][y]==0)


//if(x&&y)


if(x&&y)


g[x][y]=1;


//g[y][x]=1;


//g[i][x]=1;


//g[i][y]=1;


}


int sum=0;


sum=MaxMatch();


printf("%d\n",uN-sum);

}

return
0;

}