## The shortest path

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

There are n points on the plane, Pi(xi, yi)(1 <= i <= n), and xi < xj (i<j). You begin at P1 and visit all points then back to P1. But there is a constraint:
Before you reach the rightmost point Pn, you can only visit the points those have the bigger x-coordinate value. For example, you are at Pi now, then you can only visit Pj(j > i). When you reach Pn, the rule is changed, from now on you can only visit the points those have the smaller x-coordinate value than the point you are in now, for example, you are at Pi now, then you can only visit Pj(j < i). And in the end you back to P1 and the tour is over.
You should visit all points in this tour and you can visit every point only once.

Input

The input consists of multiple test cases. Each case begins with a line containing a positive integer n(2 <= n <= 200), means the number of points. Then following n lines each containing two positive integers Pi(xi, yi), indicating the coordinate of the i-th point in the plane.

Output

For each test case, output one line containing the shortest path to visit all the points with the rule mentioned above.The answer should accurate up to 2 decimal places.

Sample Input

3 1 1 2 3 3 1

Sample Output

6.47 Hint: The way 1 - 3 - 2 - 1 makes the shortest path.

J.L.Bentley建议通过只考虑双调旅程来简化问题，这种旅程即为从最左点开始，严格地从左到右直至最右点，然后严格地从右到左直至出发点。图b显示了同样的7个点问题的最短双调路线。在这种情况下，多项式时间的算法是可能的。

AC:

``````#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

const int inf = 0x3f3f3f3f;
int n;
struct node
{
double x,y;
}pos[205];
double dp[205][205];

double distance(int i,int j)
{
double tmp;
tmp = (pos[i].x - pos[j].x)*(pos[i].x - pos[j].x) + (pos[i].y - pos[j].y)*(pos[i].y - pos[j].y);
return sqrt(tmp);
}

int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i = 1; i <= n; i++)
scanf("%lf %lf",&pos[i].x,&pos[i].y);
dp[1][2] = distance(1,2);
for(int j = 3; j <= n; j++)
{
// i < j-1   dp[i][j] = dp[i][j-1] + distance(j-1,j);
for(int i = 1; i < j - 1; i++)
{
dp[i][j] = dp[i][j-1] + distance(j-1,j);
}
dp[j-1][j] = inf;
//i = j-1   dp[i][j] = min{dp[k][i]+distance(k,j)}
for(int i = 1; i < j - 1; i++)
{
double tmp = dp[i][j-1] + distance(i,j);
if(tmp < dp[j-1][j])
dp[j-1][j] = tmp;
}
}
dp[n][n] = dp[n-1][n] + distance(n-1,n);
printf("%.2lf\n",dp[n][n]);
}
return 0;
}``````