pku 1111 解题报告简单的广度优先搜索

最新推荐文章于 2020-01-21 22:16:29 发布

原创最新推荐文章于 2020-01-21 22:16:29 发布 · 572 阅读

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#struct

本文提供了一个简单的广度优先搜索(BFS)入门题目的代码实现，通过实例详细展示了如何使用队列来遍历二维数组，寻找可达路径并计算特定位置的数量。

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http://acm.pku.edu.cn/JudgeOnline/problem?id=1111

这个题目不难，是个广搜的入门题目

#include<iostream>
#include<queue>
using namespace std;
bool b[22][22];
char a[22][22];
struct node
{
int x;
int y;
node( int X = 0, int Y = 0 ){ x = X; y = Y; }
}no;
int gr,gc,mr,mc,i,j,sum=0;
char c;
int main()
{

while((cin>>gr>>gc>>mr>>mc)&&(gr!=0||gc!=0||mr!=0||mc!=0))
{
  memset(b,0,sizeof(b));
  memset(a,'/0',sizeof(a));
  sum=0;
  gets(a[1]+1);
  for(j=1;j<=gr;j++)
   gets(a[j]+1);
  //cout<<a;
  char w=a[mr][mc];
  b[mr][mc]=1;
  if(w=='.')
  {
   cout<<sum<<endl;
   continue;
  }
  queue<node>q;
  q.push(node(mr,mc));
  while(!q.empty())
  {
   no=q.front();
   if(a[no.x-1][no.y-1]=='X'&&b[no.x-1][no.y-1]==0)
   {
    q.push(node(no.x-1,no.y-1));
    b[no.x-1][no.y-1]=1;
   }
  // else //if(a[no.x-1][no.y-1]=='.')
   // sum++;
   if(a[no.x+1][no.y-1]=='X'&&b[no.x+1][no.y-1]==0)
   {
    q.push(node(no.x+1,no.y-1));
    b[no.x+1][no.y-1]=1;
   }
  // else if(a[no.x+1][no.y-1]=='.')
  //  sum++;
   if(a[no.x+1][no.y+1]=='X'&&b[no.x+1][no.y+1]==0)
   {
    q.push(node(no.x+1,no.y+1));
    b[no.x+1][no.y+1]=1;
   }
//  else if(a[no.x+1][no.y+1]=='.')
//   sum++;
   if(a[no.x-1][no.y+1]=='X'&&b[no.x-1][no.y+1]==0)
   {
    q.push(node(no.x-1,no.y+1));
    b[no.x-1][no.y+1]=1;
   }
//  else if(a[no.x-1][no.y+1]=='.')
//   sum++;
   if(a[no.x][no.y+1]=='X'&&b[no.x][no.y+1]==0)
   {
    q.push(node(no.x,no.y+1));
    b[no.x][no.y+1]=1;
   }
   else if(a[no.x][no.y+1]=='.'||a[no.x][no.y+1]=='/0')
    sum++;
   if(a[no.x][no.y-1]=='X'&&b[no.x][no.y-1]==0)
   {
    q.push(node(no.x,no.y-1));
    b[no.x][no.y-1]=1;
   }
   else if(a[no.x][no.y-1]=='.'||a[no.x][no.y-1]=='/0')
    sum++;
   if(a[no.x+1][no.y]=='X'&&b[no.x+1][no.y]==0)
   {
    q.push(node(no.x+1,no.y));
    b[no.x+1][no.y]=1;
   }
   else if(a[no.x+1][no.y]=='.'||a[no.x+1][no.y]=='/0')
    sum++;
   if(a[no.x-1][no.y]=='X'&&b[no.x-1][no.y]==0)
   {
    q.push(node(no.x-1,no.y));
    b[no.x-1][no.y]=1;
   }
   else if(a[no.x-1][no.y]=='.'||a[no.x-1][no.y]=='/0')
    sum++;
   q.pop();
  }
  cout<<sum<<endl;
}
return 0;
}