Tour - UVa 1347 dp

最新推荐文章于 2020-10-07 17:45:39 发布

原创最新推荐文章于 2020-10-07 17:45:39 发布 · 769 阅读

0 ·

CC 4.0 BY-SA版权

dp动态规划和递推同时被 2 个专栏收录

281 篇文章

订阅专栏

UVa

218 篇文章

订阅专栏

本文介绍了一种计算平面中给定点集的最短封闭旅行路径的方法。通过动态规划策略实现，确保路径始终沿x轴方向前进，适用于节省旅行成本的问题。

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 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 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

Sample Output

6.47
7.89

题意：按x轴排序给出点，求从最左边的点到最右边的点，再回来后路径的最小值。

思路：看成两条线从左到右，dp[i][j]表示1-max(i,j)的点都经过后的最短路径。两条路径一定都是从x小的点向x大的点运动的，返回的情况会使结果更大。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
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\n",dp[n][n]);
  }
}