 Tour - UVA - 1347-动态规划

***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

```#include

#include

#include

#include

#include

#include

#include

#include

#include

#define maxn 50005using namespace std;typedef long long ll;int n;struct F{
int x,y;};F point;double dist;double d;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);
}
return 0;}

```

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[***.191.171.42]2022年08月18日 08时35分13秒

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