Ural 1143. Electric Path

1143. Electric Pathtime limit:1.0 Second
Memory limit:64 Mbbackgroundat The team competition of the 10th National Student Informatics Olympic, which is organized At Hanoi National University, there is NTeams participating. Each team was assigned to work in a camp. On the map, it can is seen that the camps is positioned on the vertices of a convex polygon with NVertices P1, P2,..., PN(The vertices is enumerated around the polygon in counter-clockwise order.) In order to achieve absolute safety providing electricity to the camps, besides an electric supplying system, the host Org Anization set up a path from a reserved electricity generator (which was placed in one of the camps) to every camp once, an D The path ' s total length is minimum. Problemgiven the coordinates of the polygons ' vertices (the camps ' positions), determine the length of the electric path C Orresponding to the host organization ' s arrangement. Inputthe first line contains the integer N(1≤ N≤200). The I' th line of the next NLines contains, real numbers XI, Yi, separated by a space, with no more than 3 digits after the decimal points, is vertex Pi' s coordinates on the plane (with I= 1, 2, ..., N). The length of the path connecting the vertex ( XI, Yi) and ( XJ, YJ) is computed with the FORMULA:SQRT ( XI− XJ) 2 + ( Yi− YJ) 2). Outputthe only line should contain real number L(written in real number format, with 3 digits after the decimal point), which are the total length of the electric path. Sample

input	Output
450.0 1.05.0 1.00.0 0.045.0 0.0	50.211

problem Source:The competition for selecting the Vietnam IOI team Tags:None()difficulty:532 Test Instructions: The order gives the n points of a convex polygon on a plane, asking for the shortest length of a line to connect them. Analysis: First, you can feel that the line will not cross.　　Then we prove it: Consider the situation of four points first. Take four points of intersection.　　If cross, then the four-sided shape of these four points, it is certain that two diagonal lines will be selected, and then add an edge. Then, obviously, you can change a diagonal edge to an adjacent edge (connecting adjacent points).　　This is obviously shorter.　　Consider more points of the situation.　　You can select two endpoints of two lines that intersect, with a total of four points.　　The other points are then abstracted into edges, and because of the convex polygon, the same nature must be satisfied. It can be abstracted into four-point situations. Then there is this nature that can be done. Starting from both sides dp,dp[i][j][0..1] means from the first I to the first J, from the left or the right to start wiring. The transfer is obvious, there are only two kinds.

1 /**2 Create by Yzx-stupidboy3 */4#include <cstdio>5#include <cstring>6#include <cstdlib>7#include <cmath>8#include <deque>9#include <vector>Ten#include <queue> One#include <iostream> A#include <algorithm> -#include <map> -#include <Set> the#include <ctime> -#include <iomanip> - using namespacestd; -typedefLong LongLL; +typedefDoubleDB; - #defineMIT (2147483647) + #defineINF (1000000001) A #defineMLL (1000000000000000001LL) at #defineSZ (x) ((int) (x). Size ()) - #defineCLR (x, y) memset (x, y, sizeof (x)) - #definePUF Push_front - #definePub push_back - #definePOF Pop_front - #definePOB pop_back in #defineFT first - #defineSD Second to #defineMk Make_pair +  -InlineintGetint () the { *     intRet =0; $     CharCh =' ';Panax Notoginseng     BOOLFlag =0; -      while(! (Ch >='0'&& Ch <='9')) the     { +         if(Ch = ='-') Flag ^=1; ACh =GetChar (); the     } +      while(Ch >='0'&& Ch <='9') -     { $RET = RET *Ten+ Ch-'0'; $Ch =GetChar (); -     } -     returnFlag? -Ret:ret; the } - Wuyi Const intN =410; the struct Point - { Wu DB x, y; -  AboutInlinevoidRead () $     { -scanf"%LF%LF", &x, &y); -     } - } Arr[n]; A intN; +DB dp[n][n][2]; the BOOLvisit[n][n][2]; -  $InlinevoidInput () the { thescanf"%d", &n); the      for(inti =1; I <= N; i++) Arr[i]. Read (); the } -  in Inline DB SQR (db x) the { the     returnX *x; About } the  theInline DB Dist (ConstPoint &a,ConstPoint &B) the { +     returnsqrt (SQR (a.x-b.x) + SQR (A.Y-b.y)); - } the BayiInline DB Work (intLeftintRightBOOLtype) the { the     if(left >= right)return 0; -     if(Visit[left][right][type]) -         returnDp[left][right][type]; theVisit[left][right][type] =1; theDB ret =1.0*INF, CNT; the     if(Type = =0) the     { -CNT = Dist (Arr[left], Arr[left +1]); theCNT + = Work (left +1, right,0); theRET =min (ret, CNT); the 94CNT =Dist (Arr[left], arr[right]); theCNT + = Work (left +1, right,1); theRET =min (ret, CNT); the     }98     Else About     { -CNT = Dist (Arr[right], arr[right-1]);101CNT + = Work (left, right-1,1);102RET =min (ret, CNT);103 104CNT =Dist (Arr[right], arr[left]); theCNT + = Work (left, right-1,0);106RET =min (ret, CNT);107     }108 109     returnDp[left][right][type] =ret; the }111  theInlinevoidSolve ()113 { the      for(inti = n +1; I <=2N i++) theArr[i] = arr[i-n]; the 117DB ans =1.0*INF, CNT;118      for(inti =1; I <= N; i++)119     { -CNT = Work (i, i + N-1,0);121Ans =min (ans, cnt);122CNT = Work (i, i + N-1,1);123Ans =min (ans, cnt);124     } the 126printf"%.3lf\n", ans);127 } - 129 intMain () the {131Freopen ("a.in","R", stdin); the Input ();133 Solve ();134     return 0;135}

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Ural 1143. Electric Path