POJ2139 解题报告

Six Degrees of Cowvin Bacon

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 5769Accepted: 2714

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 23 1 2 32 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

这是一个简单的求 任意两点之间最短路 的问题

使用Floyd算法可以很快地解决

不过这一题由于一开始没有理解题意所以wa了

这一题的大意是，求解出一头离其他的牛距离之和最小的牛，然后算出这个距离之和的平均值再乘上100，得出来的就是答案

#include<iostream>#include<algorithm>using namespace std;const int INF=1000;const int maxn=300+30;const int maxm=10000+100;int dp[maxn][maxn];int cows[maxn];int ncow;int N,M;void set_cows();void floyd();void solve();void show();int main(){    while(cin>>N>>M)    {        fill(dp[0],dp[0]+maxn*maxn,INF);        for(int i=0;i<M;i++)        {            cin>>ncow;            for(int i=0;i<ncow;i++)                cin>>cows[i];            set_cows();        }        floyd();        //show();        solve();    }    return 0;}void set_cows(){    for(int i=0;i<ncow;i++)        for(int j=0;j<ncow;j++)            if(i!=j) dp[cows[i]][cows[j]]=dp[cows[j]][cows[i]]=1;            else dp[cows[i]][cows[j]]=dp[cows[j]][cows[i]]=0;}void floyd(){    for(int k=1;k<=N;k++)        for(int i=1;i<=N;i++)            for(int j=1;j<=N;j++)                dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);}void solve(){    int sum,min_sum=INF;    for(int i=1;i<=N;i++)    {        sum=0;        for(int j=1;j<=N;j++)            sum+=dp[i][j];        min_sum=min(sum,min_sum);    }    cout<<min_sum*100/(N-1)<<endl;}void show(){    for(int i=1;i<=N;i++)    {        for(int j=1;j<=N;j++)            if(dp[i][j]<INF) cout<<dp[i][j]<<"    ";            else cout<<"inf   ";        cout<<endl;    }}