|
Language:
Six Degrees of Cowvin Bacon
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 |
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 pf printf
#define DBG pf("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;
}
Six Degrees of Cowvin Bacon (poj 2139 最短路Floyd)
原文地址：http://blog.csdn.net/u014422052/article/details/44535539
