Challenge Program 2.1.5 Exhaustion Search >> Width First search (practice POJ3669)

Compare DFS and BFS first

Depth-First Search Dfs width first search BFS

You can clearly see that the search order is different.

DFS is to search the bottom of a single path, and then backtrack.

BFS is the search near the state, until the bottom, generally in solving the shortest path or minimum number of steps to apply.

BFS to use the queue it.

The use of queues take a look at http://blog.csdn.net/cindywry/article/details/51919282

　　

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Title: poj3669

Because of the number of steps (foldback) The problem was one day.

If that point can go, then push it into the queue and wait for the search.

1#include <stdio.h>2#include <queue>3#include <string.h>4 using namespacestd;5 intmap1[305][305],step[305][305];6typedef pair<int,int>P;7 structstu{8 intXintYintT;9 };Ten structStu ch[50002]; One  A intdx[5]={0,1,0, -1,0}; - intdy[5]={0,0,1,0, -1}; -  the voidZhaintM) { -memset (MAP1,-1,sizeof(MAP1)); -      for(inti =0; i < M; i++) -          for(intK =0; K <5; k++) {//Five Points is burst. +             intNX = ch[i].x +Dx[k]; -             intNY = ch[i].y +Dy[k]; +             if(NX >=0&& NX <303&& NY >=0&& NY <303&& (ch[i].t < Map1[ny][nx] | | map1[ny][nx] = =-1)) AMAP1[NY][NX] = ch[i].t;//give Everypoint the minimum burst time. at         } - } -  - intBFs () { -Queue <P>que; -Que.push (P (0,0)); inmemset (step,-1,sizeof(step)); -step[0][0]=0; to  while(Que.size ()) { +P p=Que.front (); - Que.pop (); the         if(Map1[p.first][p.second] = =-1){//find Map1 ' s-1 and print it *             returnStep[p.first][p.second]; $         }Panax Notoginseng     if(Step[p.first][p.second] >= Map1[p.first][p.second])Continue;//The point had Benn Burst,so jump it -      for(inti =1; I <5; i++) {//This is the point of four diretions, does not been destroy. the         intNX = P.first + Dx[i],ny = P.second +Dy[i]; +         if(nx>=0&& NY >=0&&step[nx][ny] = =-1)//if it can arrive and does not exceed the map A             { theQue.push (P (NX, NY));//Push It to the queue and go +step[nx][ny]=step[p.first][p.second]+1;//of Course,step must add 1 when go. -             } $     } $ } - return-1; - } the  - intMain ()Wuyi { the     intM,a; -      while(~SCANF ("%d",&M)) Wu     { -           for(intI=0; i<m; i++) Aboutscanf"%d%d%d",&ch[i].x,&ch[i].y,&ch[i].t); $ Zha (M); -A=BFS (); -printf"%d\n", a); -     } A     return 0; +}

Challenge Program 2.1.5 Exhaustion Search >> Width First search (practice POJ3669)