Hdu 2337 Escape from Enemy Territory

Main topic

Give you a map of nn*mm rectangles. Some of the above points have enemy camps. Give you the beginning and the end, you find an optimal path. The shortest distance from the enemy battalion that satisfies the point on the optimal path is the shortest of all paths. If there is more than one path to find the shortest.

Analysis

The shortest distance from the enemy battalion is determined by two points. Then BFS to verify.

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5#include <cmath>6#include <string>7#include <queue>8#include <stack>9#include <map>Ten#include <vector> One #defineMAXN 1010 A #defineMAXN 2005 - #defineMAXM 20000005 - #defineINF 100000000000000 the #defineOO 1000000007 - using namespacestd; -  -typedefLong LongLL; + struct Point - { +     intx,y,dist; A}p[maxn*MAXN]; at intMid,n,nn,mm,stx,sty,edx,edy; - intVIST[MAXN][MAXN],VISTDIST[MAXN][MAXN]; - intdirx[]={0,-1,0,1}; - intdiry[]={-1,0,1,0}; - intMax_dist,dist1,dist2; - voidBFS1 () in { -Queue<point>Q; to      while(!q.empty ()) + Q.pop (); -      for(intI=0; i<n;i++) the Q.push (P[i]); *      while(!q.empty ()) $     {Panax NotoginsengPoint tem=Q.front (); - Q.pop (); the  +          for(intI=0;i<4; i++) A         { the            intnewx=tem.x+Dirx[i]; +            intnewy=tem.y+Diry[i]; -            if(newx>=0&&newx<mm&&newy>=0&&newy<nn&&!Vist[newx][newy]) $            { $vist[newx][newy]=1; -                intnewdist=tem.dist+1; -vistdist[newx][newy]=newdist; themax_dist=Max (max_dist,newdist); - Point pp;Wuyipp.x=newx; thepp.y=Newy; -pp.dist=newdist; Wu Q.push (PP); -            } About         } $     } - } - intBFS2 () - { AQueue<point>Q; +      while(!q.empty ()) the Q.pop (); - Point pp; $pp.x=STX; thepp.y=sty; thepp.dist=0; the Q.push (PP); thememset (Vist,0,sizeof(vist)); -     if(vistdist[stx][sty]<mid) in         return 0; the  the      while(!q.empty ()) About     { thePoint tem=Q.front (); the Q.pop (); the         if(tem.x==edx&&tem.y==Edy) +         { -dist1=mid; theDist2=tem.dist;Bayi             return 1; the         } the          for(intI=0;i<4; i++) -         { -             intxx=tem.x+Dirx[i]; the             intyy=tem.y+Diry[i]; the             if(xx>=0&&xx<mm&&yy>=0&&yy<nn&&!vist[xx][yy]&&vistdist[xx][yy]>=mid) the             { thevist[xx][yy]=1; - Point pp; thepp.x=xx; thepp.y=yy; thepp.dist=tem.dist+1;94 Q.push (PP); the             } the  the         }98     } About     return 0; - 101 }102 intMain ()103 {104     intT; thescanf"%d",&t);106      while(t--)107     {108scanf" %d%d%d",&n,&mm,&nn);109scanf"%d %d%d%d",&stx,&sty,&edx,&edy); thememset (Vist,0,sizeof(vist));111          for(intI=0; i<n;i++) the         {113scanf"%d%d",&p[i].x,&p[i].y); thep[i].dist=0; thevistdist[p[i].x][p[i].y]=0; thevist[p[i].x][p[i].y]=1;117         }118max_dist=-1;119 BFS1 (); -         intL=0, r=max_dist;121          while(l<=R)122         {123Mid= (L+R)/2;124             if(BFS2 ()) theL=mid+1;126             Else127r=mid-1; - 129         } theprintf"%d%d\n", dist1,dist2);131     } the     return 0;133}

Note that the number of queues is small, and that is what should be written when you insert a queue.

Point      pp;               pp.x=newx;               Pp.y=newy;               Pp.dist=newdist;               Q.push (PP);

If this is written, it will be CE

Q.push (point) {newx,newy,newdist});

Hdu 2337 Escape from Enemy Territory