Language:
Six Degrees of Cowvin Bacon
Time Limit:1000MSMemory Limit:65536K
Total Submissions:3288Accepted:1529
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
题意：英语太差，读题就读了半小时n头牛，m行关系，每一行先输入头数num，紧接着num头牛，，这num头牛两两之间距离为1，最后问哪一头牛到其他所有牛距离的平均值最小，ans=到其他所有牛的最短距离之和*100/（n-1）。
代码：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 330#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b) for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i–)#define FRL(i,a,b) for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define DBGpf("Hi\n")typedef long long ll;using namespace std;int mp[maxn][maxn];int n,m;int w[maxn];void Floyd(){int i,j,k;FRE(k,1,n){FRE(i,1,n){FRE(j,1,n){if (i!=j&&i!=k&&mp[i][k]<INF)mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);}}}}int main(){int i,j,k;while (~sff(n,m)){mem(mp,INF);FRL(i,0,m){int num;sf(num);FRL(j,0,num)sf(w[j]);FRL(j,0,num){FRL(k,0,j)mp[ w[j] ][ w[k] ]=mp[ w[k] ][ w[j] ]=1;}}Floyd();int ans=INF;FRE(i,1,n){int s=0;FRE(j,1,n){if (i!=j && mp[i][j]<INF)s+=mp[i][j];}ans=s<ans?s:ans;}pf("%d\n",(int)((ans*100)/(n-1)));}return 0;}
要想捉大鱼，不能怕水深。要想摘玫瑰，就得不怕刺。
