poj2139 Six Degrees of Cowvin Bacon(Flyod)
2014-12-06 19:06
344 查看
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
Sample Output
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
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 3155 | Accepted: 1464 |
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
/* 注意不要把0给初始化掉。。无语 加油!!! Time:2014-12-6 19:06 */ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int INF=1<<20; const int MAX=300+10; int n,m;int data[MAX]; int map[MAX][MAX]; void Init(){ memset(data,0,sizeof(data)); for(int i=1;i<=n;i++){//注意,不要把0初始化掉。。。。。 for(int j=i;j<=n;j++){ if(i==j) map[i][j]=0; else map[i][j]=map[j][i]=INF; } } } void Floyd(int n){ for(int k=1;k<=n;k++){ for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ if(map[i][j]>map[i][k]+map[k][j]) map[i][j]=map[j][i]=map[i][k]+map[k][j]; } } } } int main(){ int num; while(scanf("%d%d",&n,&m)!=EOF){ Init(); while(m--){ scanf("%d",&num); for(int i=0;i<num;i++){ scanf("%d",&data[i]); for(int j=0;j<i;j++){ map[data[j]][data[i]]=map[data[i]][data[j]]=100; } } } Floyd(n); int minV=INF; for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ map[i][0]+=map[i][j]; } if(minV>map[i][0]) minV=map[i][0]; } printf("%d\n",minV/(n-1)); } return 0; }
相关文章推荐
- POJ2139 Six Degrees of Cowvin Bacon 【Floyd】
- poj2139 Six Degrees of Cowvin Bacon最短路问题
- 【Floyd】POJ2139 -Six Degrees of Cowvin Bacon
- POJ2139 Six Degrees of Cowvin Bacon 最短路
- POJ2139 Six Degrees of Cowvin Bacon
- POJ2139 Six Degrees of Cowvin Bacon
- POJ2139-Six Degrees of Cowvin Bacon
- POJ2139-Six Degrees of Cowvin Bacon
- POJ - 2139 Six Degrees of Cowvin Bacon(图论/无权最短路径BFS)
- POJ2139 Six Degrees of Cowvin Bacon [Floyd]
- poj 2139 Six Degrees of Cowvin Bacon (Floyd 算法)
- Six Degrees of Cowvin Bacon POJ - 2139 弗洛伊德算法
- POJ 2139 Six Degrees of Cowvin Bacon (Floyd)
- Six Degrees of Cowvin Bacon(最短路径floyd算法)
- POJ 2139 Six Degrees of Cowvin Bacon (Floyd)
- Six Degrees of Cowvin Bacon poj 2131(floyd)
- Six Degrees of Cowvin Bacon
- poj 2139 Six Degrees of Cowvin Bacon floyd算法
- POJ 2139 Six Degrees of Cowvin Bacon
- POJ 2139 Six Degrees of Cowvin Bacon