B-Nana in Wonderland Series-long jump Queen

B-Nana in Wonderland Series-long jump QueenTime Limit:2000/1000ms (java/others) Memory Limit: 128000/64000kb (java/others) Problem Description

Nana thought the piano is very uninteresting, abandoned the piano, continue to walk, Front is a lake, Nana thought of the other side of the lake, but Nana looked for a long time did not find the small bridge and boat, Nana also found that they are not immortal, not like eight Immortals crossing. Just when Nana worried, Nana found some rocks above the lake! Nana Brainwave, found that can follow the stone to jump acridine, so keep jumping, perhaps can jump to the other side!

Nana to the location of all the stones to tell you, and then Nana can jump the farthest distance is also know ~ please smart you tell Nana, she can reach the other side smoothly?

In order to be able to express the position of each stone smoothly, assuming that Nana is on the x-axis, the lake on the other side of the bank is a straight line y=y0, the stones in the lake are ordered two-tuple <x,y>, we can assume that the lake is infinite width, two stone distance for the geometric distance, The distance between the stone and the shore is from the point to the straight line.

Input

Multiple sets of data, first a positive integer T (t<=20) representing the number of data groups

For each set of data is first three integers y0 (1<=y0<=1000), N (0<=n<=1000), D (0<=d<=1000), respectively, the location of the lake on the other shore, the number of stones, Nana once the furthest distance to jump.

Next is n rows, each line being two integers x, y (0<=|x|<=1000,0<y<y0)

Output

For each set of data, if Nana can reach another shore of the lake, first output "YES", and then output an integer, which means Nana at least how many times to jump to the other shore,

If Nana cannot reach another shore of the lake, first output "NO", and then output an integer that represents Nana's closest distance from the other shore of the lake. (Note case)

Sample Input

24 3 10 10 20 36 3 20 11 22 3

Sample Output

Yes4no3

Hint

Example one, from the opposite side of the 0,2 (0,3), x-axis (0,1), a total of 4 steps, Output 4

Example two, from the x-axis (0,1)---(2,3), at this time the distance from the other shore is 3, the maximum jump distance of 2, can not reach the other side, so output 3

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6#include <string>7#include <vector>8#include <stack>9#include <queue>Ten#include <Set> One#include <map> A#include <list> -#include <iomanip> -#include <cstdlib> the#include <sstream> - using namespacestd; -typedefLong LongLL; - Const intinf=0x5fffffff; + Const Doubleexp=1e-6; - Const intms=1005; +  A intDES,N,MAXV; at  -  - DoubleEdges[ms][ms]; - intCnt[ms]; - BOOLFlag[ms]; -  in  - struct Point to { +       intx, y; - }points[ms]; the  * DoubleDistence (Point &a,point &b) $ {Panax Notoginseng       returnsqrt ((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (a.y-b.y)); - } the  + intMain () A { the       intT; +scanf"%d",&T); -        while(t--) $       { $scanf"%d%d%d",&des,&n,&MAXV); -             intx, y; -              for(intI=1; i<=n;i++) the             { -scanf"%d%d",&x,&y);Wuyipoints[i].x=x; thepoints[i].y=y; -             } Wu  -              for(intI=1; i<=n;i++) Aboutcnt[i]=INF; $  -memset (Flag,false,sizeof(flag)); -  -queue<int>que; A  +              for(intI=1; i<=n;i++) the             { -                   if(points[i].y<=MAXV) $                   { thecnt[i]=1; the Que.push (i); the                   } the                   if(des-points[i].y<=MAXV) -flag[i]=true; in                    for(intj=i+1; j<=n;j++) the                   { theedges[i][j]=edges[j][i]=distence (Points[i],points[j]); About                   } the             } the              while(!que.empty ()) the             { +                   intu=Que.front (); - Que.pop (); the                    for(intv=1; v<=n;v++)Bayi                   { the                         if((u!=v) &&edges[u][v]<= (Double) (maxv+EXP)) the                               { -                                           if(cnt[u]+1<Cnt[v]) -                                           { thecnt[v]=cnt[u]+1; the Que.push (v); the                                           } the                               } -                   } the  the             } the 94             if(des<=MAXV) the             { theprintf"yes\n1\n"); the                   Continue;98             } About  -             if(n==0)101             {102                   if(des<=MAXV)103printf"yes\n1\n");104                   Else theprintf"no\n%d\n", des);106                   Continue;107             }108 109             intcn1=INF; the             intt=0;111              for(intI=1; i<=n;i++) the             {113                   if(Flag[i]) the                   { the                        if(cnt[i]<INF) the                         {117                               if(ans>cnt[i]+1)118ans=cnt[i]+1;119                         } -                   }121                  if(cnt[i]<INF)122                   {123                         if(points[i].y>t)124t=points[i].y; the                   }126             }127  -             if(ans<INF)129             { theprintf"yes\n");131printf"%d\n", ans); the             }133             Else134             {135printf"no\n");136printf"%d\n", des-t);137             }138       }139       return 0; $}

B-Nana in Wonderland Series-long jump Queen