Ultraviolet A 1347 (POJ 2677) Tour (dual-tuning Euclidean Traveling Salesman Problem)

Ultraviolet A 1347 (POJ 2677) Tour (dual-tuning Euclidean Traveling Salesman Problem)

Tour

Time Limit:3000 MSMemory Limit:0 KB64bit IO Format:% Lld & % llu

Description

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 planePI = <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 distinctX-Coordinates.

Write a program that, given a setNPoints 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 participant set of points. for each set of points the data set contains the number of points, and the point coordinates in ascending order ofXCoordinate. White spaces can occur freely in input. The input data are correct.

Output

For each set of data, your program shocould 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 theirXAndYCoordinates. The second point, for example, hasXCoordinate 2, andYCoordinate 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 12 33 14 1 1 2 33 14 2

 

Sample Output

6.477.89

 

Question: a typical dynamic planning example. It is also called the dual-tuning Euclidean Traveling Salesman Problem. Introduction to algorithms.

 

Ideas:

Dp [I] [j] indicates the distance from I to 1, and then from 1 to j. In this path, all vertices from Point 1 to Pmax (I, j) have and pass through only once.

 

Dp [I] [j] = dp [I-1] [j] + dis (I, I-1 );

Dp [I] [I-1] = min (dp [I] [I-1], dp [I-1] [j] + dis (I, j ));

 

 

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        using namespace std;const int INF = 1<<29;const int MAXN = 1100;const double PI = acos(-1.0);const double e = 2.718281828459;const double eps = 1e-8;struct node{ double x; double y;}a[MAXN];double dp[MAXN][MAXN];int cmp(node a, node b){ return a.x < b.x;}double dist(int i, int j){ return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));}int main(){ //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int n; while(cin>>n) { for(int i = 1; i <= n; i++) { scanf("%lf %lf", &a[i].x, &a[i].y); } sort(a+1, a+1+n, cmp); dp[2][1] = dist(1, 2); for(int i = 3; i <= n; i++) { dp[i][i-1] = INF*1.0; for(int j = 1; j < i-1; j++) { dp[i][i-1] = min(dp[i][i-1], dp[i-1][j]+dist(i, j)); dp[i][j] = dp[i-1][j]+dist(i-1, i); } } double ans = INF*1.0; for(int i = 1; i < n; i++) { ans = min(ans, dp[n][i]+dist(n, i)); } printf("%.2f\n", ans); } return 0;}