UVA 1347（POJ 2677) Tour(双色欧几里德旅行商问题）

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 pi = < xi,yi >. 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 distinct
x-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. 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

算是经典问题了吧。题目大意：给定平面上n个点的坐标（按x的升序），

你的任务是设计一条路线从最左边出发，到最右边再返回最左边，每点恰好经过一次。问最小代价。能够看成两个人从最左端同一时候出发。

每点恰有一人经过的最小代价。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<cmath>
typedef long long LL;
using namespace std;
const int maxn=110;
double x[maxn],y[maxn];
double dp[maxn][maxn];

double dis(int i,int j)
{
    return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}

int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=0;i<n;i++)
            scanf("%lf%lf",&x[i],&y[i]);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
                dp[i][j]=INT_MAX;
        }
        dp[0][0]=0.0;//初始化
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<i;j++)
            {
               dp[i][i-1]=min(dp[i][i-1],dp[i-1][j]+dis(i,j));//当走到dp[i-1][j]要走到第i个点时。有两种途径
               dp[i][j]=min(dp[i][j],dp[i-1][j]+dis(i-1,i));//在i-1的人走到i或在j的人走到i；
            }
        }
//        for(int i=0;i<n;i++)
//            printf("fuck %d\n",dp[n-1][i]);
        double ans=dp[n-1][n-2]+dis(n-1,n-2);
        printf("%.2f\n",ans);
    }
    return 0;
}