poj--2139
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.
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.
* 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
题意：奶牛们要拍电影，如果两个奶牛同时拍一部那么他们的距离为1，如果两个奶牛同时和第三头奶牛拍一部电影，那么他们的距离为2，求他们最短距离乘上100的最小平均值
这题是相当于求最短路径，forld最小环从自身出发，最后到达自身的最短路径，
Flord 最小环问题相当于一个动态规划的算法，如果a不能到达b那么可以考虑引入一个k，经过k到达b并更新一下a，b的距离
所以这个算法的核心代码是
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
map[i][j]=min(map[i][j],map[i][k]+map[k][j])
最近做题一直忘记预处理，导致程序不对。
#include<iostream> #include<cstring> #include<algorithm> using namespace std; const int Max_n=305; const int INF=99999999; int map[Max_n][Max_n]; int node[Max_n]; int n,m; void prepare() { for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { if(i!=j) map[i][j]=map[j][i]=INF; } } int main() { while(cin>>n>>m) { prepare(); int x; while(m--) { cin>>x; for(int i=1;i<=x;i++) { cin>>node[i]; } for(int i=1;i<=x;i++) for(int j=i+1;j<=x;j++) map[node[i]][node[j]]=map[node[j]][node[i]]=1; for(int k=1;k<=n;k++) for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) map[i][j]=min(map[i][j],map[i][k]+map[k][j]); } int ans=INF; for(int i=1;i<=n;i++) { int maxn=0; for(int j=1;j<=n;j++) { maxn+=map[i][j]; } ans=min(ans,maxn*100/(n-1)); } printf("%d\n",ans); } return 0; }
