POJ 3613
1 //共T條路，求從S到E經過N條路徑的最短路徑
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;//少加個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)//貌似運算符重載，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
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