2602 Shortest Path problem

2602 Shortest Path problem

time limit: 1 sspace limit: 32000 KBtitle level: Golden Gold SolvingTitle Description Description

There are n dots (n<=100) on the plane, and the coordinates of each point are between -10000~10000. Some of these points have connections between them. If there is a connection, it means that it is possible to reach another point from one point, that is, a path between two points and a straight distance between two points. The task now is to find the shortest path from one point to another.

Enter a description Input Description

The first behavior is an integer n.

Line 2nd to line n+1 (total n rows), two integers x and y per line, describes the coordinates of a point.

The n+2 acts as an integer m, which indicates the number of lines in the graph.

Thereafter the M-line, each line describes a connection, consisting of two integers I and j, which indicates that there is a connection between the point I and the J-point.

Last line: Two integers s and T, representing the source and target points, respectively.

Output description Output Description

Only one row, a real number (reserved two decimal places), representing the shortest path length from S to T.

Sample input Sample Input

5

0 0

2 0

2 2

0 2

3 1

5

1 2

1 3

1 4

2 5

3 5

1 5

Sample output Sample Output

3.41

Hit a water problem.

1#include <cstdio>2#include <cstdlib>3#include <cstring>4#include <cmath>5#include <iostream>6 using namespacestd;7 #defineN 1018 #defineB (i) i*i9 DoubleDis[n][n];Ten structnode{ One     intx, y; A}q[ -*N]; - DoubleRunintIintj) { -     intW=abs (Max (q[i].x,q[j].x)-min (q[i].x,q[j].x)); the     intH=abs (Max (Q[I].Y,Q[J].Y)-min (q[i].y,q[j].y)); -     returnsqrt (b (h) +B (w)); - }  - intMain () { +     intNm from, to; -scanf"%d",&n); +      for(intI=1; i<=n;i++){ Ascanf"%d%d",&q[i].x,&q[i].y); at     } -scanf"%d",&m); -memset (DIS,0x7f,sizeofdis); -      for(intI=1; i<=m;i++){ -scanf"%d%d",& from,&to ); -dis[ from][to]=dis[to][ from]=run ( from, to); in     } -      for(intk=1; k<=n;k++) to        for(intI=1; i<=n;i++) +          for(intj=1; j<=n;j++) -           if(dis[i][j]>dis[i][k]+Dis[k][j]) thedis[i][j]=dis[i][k]+Dis[k][j]; *scanf"%d%d",& from,&to ); $printf"%.2lf\n", dis[ from][to]); Panax Notoginseng     return 0; -}

2602 Shortest Path problem