首页 > 代码库 > POJ 2139 Six Degrees of Cowvin Bacon

POJ 2139 Six Degrees of Cowvin Bacon

2024-07-15 11:36:35   223人阅读

Six Degrees of Cowvin Bacon
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 2986 Accepted: 1390

Description

The cows have been making movies lately, so they are ready to play a variant of the famous game "Six Degrees of Kevin Bacon".

The game works like this: each cow is considered to be zero degrees of separation (degrees) away from herself. If two distinct cows have been in a movie together, each is considered to be one ‘degree‘ away from the other. If a two cows have never worked together but have both worked with a third cow, they are considered to be two ‘degrees‘ away from each other (counted as: one degree to the cow they‘ve worked with and one more to the other cow). This scales to the general case.

The N (2 <= N <= 300) cows are interested in figuring out which cow has the smallest average degree of separation from all the other cows. excluding herself of course. The cows have made M (1 <= M <= 10000) movies and it is guaranteed that some relationship path exists between every pair of cows.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..M+1: Each input line contains a set of two or more space-separated integers that describes the cows appearing in a single movie. The first integer is the number of cows participating in the described movie, (e.g., Mi); the subsequent Mi integers tell which cows were.

Output

* Line 1: A single integer that is 100 times the shortest mean degree of separation of any of the cows.

Sample Input

4 2
3 1 2 3
2 3 4

Sample Output

100

Hint

[Cow 3 has worked with all the other cows and thus has degrees of separation: 1, 1, and 1 -- a mean of 1.00 .]

Source

USACO 2003 March Orange


题意：牛跟自己的分离度是0，如果两牛合作分离度则为1，如果两牛同时和第三头牛合作分离度为2.求一头牛到其他牛最小的平均分离度，即求最短路。

思路：用floyd算法求最短路，注意最后结果的一个坑 要*100。


#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stdlib.h>

using namespace std;

const int INF=1000000009;
int dis[605][605];
int a[605];
int n;

void floyd(){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            for(int k=1;k<=n;k++){
                    dis[j][k]=min(dis[j][k],dis[j][i]+dis[i][k]);
            }
        }
    }
}

int main(){
    int m;
    scanf("%d%d",&n,&m);
    memset(dis,INF,sizeof(dis));
    for(int i=0;i<m;i++){
        int nn;
        scanf("%d",&nn);
        for(int j=0;j<nn;j++){
            scanf("%d",&a[j]);
        }
        for(int j=0;j<nn;j++){
            for(int k=0;k<j;k++){
                dis[a[j]][a[k]]=dis[a[k]][a[j]]=1;
            }
        }
    }
    floyd();
    int ans=INF;
    for(int i=1;i<=n;i++){
        int sum=0;
        for(int j=1;j<=n;j++){
            if(i!=j)
                sum+=dis[i][j];
        }
        ans=min(sum,ans);
    }
    printf("%d\n",int(ans*100/(n-1)));
    return 0;
}




联系
我们