Six Degrees of Cowvin Bacon POJ - 2139 (Floyd-warshall算法模板)

淩亂°似流年 2022-02-22 02:57 63阅读 0赞

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

解题思路:

利用该算法可以得出任意两点间的最短路径,最后逐个累加求出最小的平均值

注意初始化时,要把结点自己到自己的权值赋为0,到其他结点的权值赋为0x3f

#include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <stdio.h>
    using namespace std;
    const int MAXN = 310;
    const int INF = 0x3fffffff;
    int N, M;
    int dmap[MAXN][MAXN];  
    
    void warShall_flody() {
    	for (int k = 0; k < N; ++k) {
    		for (int i = 0; i < N; ++i) {
    			for (int j = 0; j < N; ++j) {
    				dmap[i][j] = min(dmap[i][j], dmap[i][k] + dmap[k][j]);
    			}
    		}
    	}
    }
    
    int main() {
    	scanf("%d %d", &N, &M);
    	memset(dmap, 0x3f, sizeof(dmap));    //必须赋值赋成0x3f
    	for (int i = 0; i < N; ++i) dmap[i][i] = 0;
    	int nums, snode;
    	for (int i = 0; i < M; ++i) {
    		scanf("%d", &nums);
    		vector<int> tnodes;
    		for (int j = 0; j < nums; ++j) {
    			scanf("%d", &snode);
    			tnodes.push_back(snode-1);  //把边插入
    		}
    		//插入到邻接矩阵中
    
    		for (int k = 0; k < tnodes.size()-1; ++k) {
    			for (int v = k + 1; v < tnodes.size(); ++v) {
    				dmap[tnodes[k]][tnodes[v]] = dmap[tnodes[v]][tnodes[k]] = 1;  //把边权设为1
    			}
    		}
    	}
    	warShall_flody();  //算出两点之间的最小值
    	int minAve =  INF;
    	for (int i = 0; i < N; ++i) {
    		int tempMin = 0;
    		for (int j = 0; j < N; ++j) {
    			tempMin += dmap[i][j];
    		}
    		minAve = min(minAve, tempMin);
    	}
    	printf("%d\n", minAve*100/(N-1));
    
    	system("PAUSE");
    	return 0;
    }