六度分离

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2927    Accepted Submission(s): 1127

【题目链接】http://acm.hdu.edu.cn/showproblem.php?pid=1869

【解题思路】Floyd算法，要证明两个素不相识的人通过中间的六个人就能联系起来，这就意味着这两个素不相识的人之间的最短距离不超过7，用Floyd算出每两个人之间的距离，然后在素不相识的情况下判断两个人之间的距离是否小于等于7，否则就不满足“六度分离”理论

 #include <iostream>
 #include <cstdio>
 #include <string>
 #include <cstring>
 #include <map>
 #include <algorithm>
 #include <cmath> #define NV 101
 #define NE 202 using namespace std; const int INF = <<;
 const double eps = 1e-;
 int ne, nv;
 int Rate[NV][NV];
 bool con[NV][NV];
 map<string, int> simap; bool Floyd()
 {
     for(int k = ; k < nv; ++k)
     for(int i = ; i < nv; ++i)
         if(Rate[i][k] < INF)
         for(int j = ; j < nv; ++j)
             if(Rate[k][j] < INF && Rate[i][j] > Rate[i][k] + Rate[k][j])
             Rate[i][j] = Rate[i][k] + Rate[k][j];
         for(int i = ; i < nv; ++i)
         for(int j = i; j < nv; ++j)
         {
             if(i != j && !con[i][j] && Rate[i][j] > )
             return false;
         }
     return true;
 } int main()
 {
     #ifndef ONLINE_JUDGE
     freopen("input.txt", "r", stdin);
     #endif
     while(cin >> nv >> ne)
     {
         for(int i=; i<nv; ++i)
         for(int j=i; j<nv; ++j)
         {
             Rate[i][j] = Rate[j][i] = INF;
             con[i][j] = con[j][i] = false;
         }         for(int i=, u, v; i<ne; ++i)
         {
             cin >> u >> v;
             Rate[u][v] = Rate[v][u] = ;
             con[u][v] = con[v][u] = true;
         }
         if(Floyd()) printf("Yes\n");
         else printf("No\n");
     }
     return ;
 }