uva1347 tour

John Doe, a skilled pilot, enjoys traveling. While on vacation, he rents a small plane and starts visiting beautiful places. To save money, John must determine the shortest closed tour that connects his destinations. Each destination is represented by a point in the plane p_i = < x_i, y_i > . John uses the following strategy: he starts from the leftmost point, then he goes strictly left to right to the rightmost point, and then he goes strictly right back to the starting point. It is known that the points have distinctx -coordinates.

Write a program that, given a set of n points in the plane, computes the shortest closed tour that connects the points according to John's strategy.

Input 

The program input is from a text file. Each data set in the file stands for a particular set of points. For each set of points the data set contains the number of points, and the point coordinates in ascending order of the x coordinate. White spaces can occur freely in input. The input data are correct.

Output 

For each set of data, your program should print the result to the standard output from the beginning of a line. The tour length, a floating-point number with two fractional digits, represents the result.

Note: An input/output sample is in the table below. Here there are two data sets. The first one contains 3 points specified by their x and y coordinates. The second point, for example, has the x coordinate 2, and the y coordinate 3. The result for each data set is the tour length, (6.47 for the first data set in the given example).

Sample Input 

3 
1 1
2 3
3 1
4 
1 1 
2 3
3 1
4 2

Sample Output 

6.47
7.89

题目大意：给出坐标平面上n个点，确定一条连接各点的最短闭合旅程的问题。

我们可以将单人旅途想象成从最左端的起点出发的两人走两条不同的路径最后在终点汇合。

动归：dp[i][j]表示一人走到i另一人走到j点时走的最短路程，因为dp[i][j]==dp[j][i]，所以为了状态的唯一，规定i>j,即状态转移方程为:dp[i][j]=min(dp[i+1][j]+dis(i,i+1)，dp[i+1][i]+dis(j,i+1)）.

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
double g[1010][1010];
double dp[1010][1010];
struct node
{
    double x,y;
} no[1010];
double dis(node &x,node &y)
{
    return sqrt((x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y));
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
               for(int i=1; i<=n; i++)
               scanf("%lf%lf",&no[i].x,&no[i].y);
               for(int i = 2; i <= n; i++)
        {
            for(int j = 1; j < i; j++)
                {
                    g[i][j] = g[j][i] = dis(no[i], no[j]);
                }
            }
        for(int i=n-1;i>=2;i--)
                for(int j=1;j