Ultraviolet A 1347 Tour DP, uva1347tourdp

Ultraviolet A 1347 Tour DP, uva1347tourdp

DP

Dp [I] [j] All the previous I points have passed. The backward person is on the j point.

Transfer dp [I + 1] [I] or dp [I + 1] [j] To point I + 1

Tour

Time Limit:3000 MS

Memory Limit:Unknown

64bit IO Format:% Lld & % llu

Submit Status

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

Source

Root: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim): Rare Topics: Rare Problems: Bitonic TSP
Root: aoapc ii: Beginning Algorithm Contests (Second Edition) (Rujia Liu): Chapter 9. Dynamic Programming: Examples
Root: aoapc I: Beginning Algorithm Contests -- Training Guide (Rujia Liu): Chapter 1. Algorithm Design: Dynamic Programming: Exercises: Intermediate

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/*************************************** * ******** Author: CKbossCreated Time:, Saturday, February 07, 2015 File Name: UVA1347.cpp *************************************** * ********/# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <string> # include <cmath> # include <cstdlib> # include <vector> # include <queue> # include <set> # include <map> using namespace std; const int maxn = 1111; Int n; struct point {double x, y;} pt [maxn]; double dist (int a, int B) {double dx = fabs (pt [a]. x-pt [B]. x); double dy = fabs (pt [a]. y-pt [B]. y); return sqrt (dx * dx + dy * dy);} double dp [maxn] [maxn]; int main () {// freopen ("in.txt ", "r", stdin); // freopen ("out.txt", "w", stdout); while (scanf ("% d", & n )! = EOF) {for (int I = 0; I <n; I ++) scanf ("% lf", & pt [I]. x, & pt [I]. y); for (int I = 0; I <maxn; I ++) for (int j = 0; j <maxn; j ++) dp [I] [j] = (1LL <60); dp [0] [0] = 0; dp [1] [0] = dist (1, 0 ); for (int I = 1; I <n; I ++) {for (int j = 0; j <I; j ++) {dp [I + 1] [I] = min (dp [I + 1] [I], dp [I] [j] + dist (j, I + 1); dp [I + 1] [j] = min (dp [I + 1] [j], dp [I] [j] + dist (I, I + 1) ;}} double ans = 1LL <60; for (int I = 0; I <n; I ++) ans = min (ans, dp [n-1] [I] + dist (I, n-1); printf ("%. 2lf \ n ", ans);} return 0 ;}