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***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 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. 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

题意：按照横坐标从左到右给出n个点，设计一条路线，从最左点出发到最右点后再返回，除最左点和最右点外，每个点只经过一次，求最短路径长度

解题思路：参考《算法竞赛入门经典》P269

#include
   
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            #define maxn 50005using namespace std;typedef long long ll;int n;struct F{ int x,y;};F point[1005];double dist[1005][1005];double d[1005][1005];double dp(int i,int j){ if(d[i][j]) return d[i][j]; if(i==n-1) return dist[n-1][n]+dist[j][n]; else return d[i][j]=min(dp(i+1,j)+dist[i][i+1],dp(i+1,i)+dist[j][i+1]);//记得保存在数组中，否则会超时}int main(){ while(scanf("%d",&n)==1) { memset(d,0,sizeof(d)); for(int i=1;i<=n;i++) scanf("%d%d",&point[i].x,&point[i].y); for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++) dist[i][j]=dist[j][i]=sqrt((point[i].x-point[j].x)*(point[i].x-point[j].x)+(point[i].y-point[j].y)*(point[i].y-point[j].y)); printf("%.2f\n",dp(2,1)+dist[1][2]); } return 0;}
           
          
         
        
       
      
     
    
   

不过我有一个疑问，按照书中的思路，点1到2的边一定经过，为什么该题的最短路径一定包含点1到2的边？

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