POJ 3660 Floyd +传递闭包问题

http://poj.org/problem?id=3660

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 54 34 23 21 22 5

Sample Output

2

有n头牛，编号为1,2……n。其中厉害的牛要排在前面，输入Ａ->Ｂ,则A比B厉害。求有几个牛的位置可以确定。

分析：如果有x头牛可以打败它，它可以打败y头牛，若x+y=n-1,则这头牛的位置可以唯一确定。

思路一：用floyd算发出每两个点之间的距离，最后统计时，若dis[a][b]之间无路且dis[b][a]之间无路，则该点位置不能确定。最后用点个数减去不能确定点的个数即可。

代码如下：

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;const int N=110,MAX=1e9+1;int dis[N][N];int n,m;void floyd(){    for(int k=1;k<=n;++k)        for(int i=1;i<=n;i++)           for(int j=1;j<=n;j++)               dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);}int main(){    while(~scanf("%d%d",&n,&m))    {        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)               dis[i][j]=MAX;        int x,y;        for(int i=1;i<=m;i++)        {           scanf("%d%d",&x,&y);           dis[x][y]=1;        }        floyd();        int ans=0;        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)            {                if(i==j)                    continue;                if(dis[i][j]==MAX&&dis[j][i]==MAX)                {                    ans++;                    break;                }            }            printf("%d\n",n-ans);    }    return 0;}

方法二：

抽象为简单的floyd传递闭包算法，在加上每个顶点的出度与入度 （出度+入度=顶点数-1，则能够确定其编号）。

传递闭包的定义：G的传递闭包定义为G*=(V,E*),其中E={(i,j):图G中存在一条从i到j的路径}。

注：有关传递闭包问题可以看这里：http://www.cnblogs.com/lpshou/archive/2012/04/27/2473109.html

代码：

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;const int maxnum=101;int map[maxnum][maxnum];int n,m;void floyd(){    int i,j,k;    for(k=1;k<=n;k++)        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)                map[i][j]=map[i][j] || (map[i][k]&&map[k][j]);//判断i和j是否连通   /* for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)            printf("%d ",map[i][j]);        printf("\n");    }*/}int main(){    scanf("%d%d",&n,&m);    memset(map,0,sizeof(map));    int i,j,u,v;    for(i=0;i<m;i++)    {        scanf("%d%d",&u,&v);        map[u][v]=1;    }    /*for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)            printf("%d ",map[i][j]);        printf("\n");    }    printf("\n");*/    floyd();    int ans,res=0;    for(i=1;i<=n;i++)    {        ans=0;        for(j=1;j<=n;j++)        {            if(i==j)    continue;            else if(map[i][j] || map[j][i])                ans++;        }        if(ans==n-1)//出度+入度=n-1            res++;    }    printf("%d\n",res);    return 0;}