poj2137 dp

 1 //Accepted    228K    32MS
 2 //dp[k][i][j] 表示从1的k号节点到i的j号节点的最小花费
 3 //dp[k][i][j]=min(dp[k][i-1][h]+cost) cost为i的j号节点与i-1的h号节点之间的距离
 4 //ans=min(dp[k][n][i]+cost) cost 为n的i号节点与1的k号节点之间的距离
 5 //事实上，数组的k这一维可以省略
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <cmath>
 9 #include <iostream>
10 using namespace std;
11 const int imax_n = 105;
12 const int imax_m = 45;
13 const double inf = 100000000.0;
14 struct Point
15 {
16     int x,y;
17 }p[imax_n][imax_m];
18 int count[imax_n];
19 double dp[imax_n][imax_m];
20 int n;
21 double min(double a,double b)
22 {
23     if (a-b<1e-9) return a;
24     return b;
25 }
26 double getCost(int i,int x,int j,int y)
27 {
28     return sqrt((double )((p[i][x].x-p[j][y].x)*(p[i][x].x-p[j][y].x)+(p[i][x].y-p[j][y].y)*(p[i][x].y-p[j][y].y)));
29 }
30 void Dp()
31 {
32     double ans=inf;
33     for (int k=1;k<=count[1];k++)
34     {
35         memset(dp,0,sizeof(dp));
36         for (int i=1;i<=n;i++)
37         for (int j=1;j<=count[i];j++)
38         dp[i][j]=inf;
39         dp[1][k]=0;
40         for (int i=2;i<=n;i++)
41         {
42             for (int j=1;j<=count[i];j++)
43             {
44                 //dp[i][j]=inf;
45                 for (int h=1;h<=count[i-1];h++)
46                 {
47                     dp[i][j]=min(dp[i][j],dp[i-1][h]+getCost(i-1,h,i,j));
48                 }
49             }
50         }
51         for (int i=1;i<=count[n];i++)
52         ans=min(ans,dp[n][i]+getCost(n,i,1,k));
53     }
54     printf("%d\n",(int )(ans*100));
55 }
56 int main()
57 {
58     scanf("%d",&n);
59     for (int i=1;i<=n;i++)
60     {
61         scanf("%d",&count[i]);
62         for (int j=1;j<=count[i];j++)
63         {
64             scanf("%d%d",&p[i][j].x,&p[i][j].y);
65         }
66     }
67     Dp();
68     return 0;
69 }