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/*
解决此题，关键在于构图。
目标：
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.
即：求解过任意两点间的最短路后，sumMin = min{ sum{dp[1][1->n]}, sum{dp[2][1->n]} ... , sum{dp[n][1->n]} }
ans = sumMin*100/(n-1)
注意：一定要先乘以100，再除以(n-1)，因为这个wa了好多次。
构图的基础：
The cows have made M (1 <= M <= 10000) movies and it is guaranteed that some relationship
path exists between every pair of cows.
即： dp[cows[Mi]][cows[Mj]] = dp[cows[Mj]][cows[Mi]] = 1 (Mi != Mj)
dp[cows[Mi]][cows[Mj]] = 0 (Mi == Mj)
*/
1 #include <iostream> 2 #include <cstdlib> 3 #include <cstdio> 4 #include <cstddef> 5 #include <iterator> 6 #include <algorithm> 7 #include <string> 8 #include <locale> 9 #include <cmath> 10 #include <vector> 11 #include <cstring> 12 #include <map> 13 #include <utility> 14 #include <queue> 15 #include <stack> 16 #include <set> 17 #include <functional> 18 using namespace std; 19 const int INF = 0x3f3f3f3f; 20 const int MaxN = 305; 21 const int MaxM = 10010; 22 const int modPrime = 3046721; 23 24 int n, m; 25 int dp[MaxN][MaxN]; 26 int cows[MaxM]; 27 28 void Solve() 29 { 30 for (int k = 1; k <= n; ++k) 31 { 32 for (int i = 1; i <= n; ++i) 33 { 34 for (int j = 1; j <= n; ++j) 35 { 36 dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]); 37 } 38 } 39 } 40 int sumMin = INF; 41 for (int i = 1; i <= n; ++i) 42 { 43 int sum = 0; 44 for (int j = 1; j <= n; ++j) 45 { 46 sum += dp[i][j]; 47 } 48 sumMin = min(sumMin, sum); 49 } 50 printf("%d\n", sumMin*100/(n-1)); 51 } 52 53 int main() 54 { 55 #ifdef HOME 56 freopen("in", "r", stdin); 57 //freopen("out", "w", stdout); 58 #endif 59 60 scanf("%d %d", &n, &m); 61 for (int i = 1; i <= n; ++i) 62 { 63 for (int j = 1; j <= n; ++j) 64 { 65 if (i != j) 66 { 67 dp[i][j] = INF; 68 } 69 else 70 { 71 dp[i][j] = 0; 72 } 73 } 74 } 75 for (int i = 0; i < m; ++i) 76 { 77 int numOfCows; 78 scanf("%d", &numOfCows); 79 for (int j = 0; j < numOfCows; ++j) 80 { 81 scanf("%d", &cows[j]); 82 } 83 for (int j = 0; j < numOfCows; ++j) 84 { 85 for (int k = j + 1; k < numOfCows; ++k) 86 { 87 dp[cows[j]][cows[k]] = dp[cows[k]][cows[j]] = 1; 88 } 89 } 90 } 91 Solve(); 92 93 #ifdef HOME 94 cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl; 95 _CrtDumpMemoryLeaks(); 96 #endif 97 return 0; 98 }
最短路（构图） 之 poj 2139 Six Degrees of Cowvin Bacon
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原文地址：http://www.cnblogs.com/shijianming/p/5027317.html
