最短路（构图） 之 poj 2139 Six Degrees of Cowvin Bacon

/*
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 }