Tour - UVa 1347 dp

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 ycoordinate 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

제목: x축에 따라 정렬하여 점을 주고 가장 왼쪽에서 가장 오른쪽까지의 점을 구하고 다시 돌아온 경로의 최소값을 구한다.
사고방식: 두 선이 왼쪽에서 오른쪽으로 보이고 dp[i][j]는 1-max(i, j)의 점이 모두 지나간 후의 가장 짧은 경로를 나타낸다.두 경로는 반드시 x의 작은 점에서 x의 큰 점으로 운동하는 것이기 때문에 되돌아오는 상황은 결과를 더욱 크게 할 것이다.
AC 코드는 다음과 같습니다.

#include
#include
#include
#include
using namespace std;
double dp[1010][1010],INF=1000000000;
int x[1010],y[1010];
double dis(int a,int b)
{ return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));
}
int main()
{ int n,i,j,k;
  while(~scanf("%d",&n))
  { for(i=1;i<=n;i++)
     scanf("%d%d",&x[i],&y[i]);
    for(i=1;i<=n;i++)
     for(j=1;j<=n;j++)
      dp[i][j]=INF;
    dp[1][1]=0;
    for(i=1;i<=n;i++)
     for(j=i;j<=n;j++)
     { dp[j][j]=min(dp[j][j],dp[i][j]+dis(i,j));
       dp[i][j+1]=min(dp[i][j+1],dp[i][j]+dis(j,j+1));
       dp[j][j+1]=min(dp[j][j+1],dp[i][j]+dis(i,j+1));
     }
    printf("%.2f
",dp[n][n]);
  }
}