POJ 3613 

 1 //共T条路，求从S到E经过N条路径的最短路径 
 2 #include <iostream>
 3 #include <fstream>
 4 #include <string.h>
 5 #include <algorithm>
 6 using namespace std;
 7 
 8 #define CIR(n, m) for (int i=0; i<n; i++) for (int j=0; j<m; j++)
 9 const int INF = 0x3fffffff;//少加个3就wa 
10 
11 struct Node
12 {
13     Node()
14     {
15           CIR(301,301) a[i][j]=INF;
16     }
17     int a[301][301];
18     int* operator [](int x)//貌似运算符重载，可惜不懂啊  
19     {
20           return a[x];
21     }
22 };
23 
24 int N, T, S, E;
25 int f[10001] = {0}, top=0;
26 Node map;
27 
28 void floyd(Node &a, Node &b, Node &c)
29 {
30     Node temp;
31     for (int k=1; k<=top; k++)
32         for (int i=1; i<=top; i++)
33             for (int j=1; j<=top; j++)
34                 temp[i][j] = min(temp[i][j], a[i][k]+b[k][j]);
35     CIR(301,301) c[i][j] = temp[i][j];
36 }
37 
38 int find(int num)
39 {
40     Node ans;
41     for (int i=0; i<301; i++) ans[i][i]=0;
42     while (num)
43     {
44         if (num&1) 
45           floyd(map, ans, ans);
46         floyd(map, map, map);
47         num >>= 1;
48     }
49     return ans[f[S]][f[E]];
50 }
51 
52 int main()
53 {
54     int i,j,k;
55     int length, from, to;
56     cin>>N>>T>>S>>E;
57     for (i=0; i<T; i++)
58     {
59         cin>>length>>from>>to;
60         if (!f[from]) f[from]=++top;
61         if (!f[to]) f[to]=++top;
62         map[f[from]][f[to]] = map[f[to]][f[from]] = length;
63     }  
64     cout<<find(N)<<endl;
65     return 0;
66 }


不懂啊     不懂啊

posted @ 2012-08-30 14:16  加拿大小哥哥  阅读(620)  评论(0)  编辑  收藏  举报