poj2243

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题目链接：

http://poj.org/problem?id=2243

题目：

Description

Input

Output

Sample Input

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6

Sample Output

To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.

Source

我用两种方法写了一下。。。

bfs版本：

#include<cstdio>
#include<queue>
#include <vector>
#include<iostream>
#include<cstring>
#include<algorithm>
const int maxn=10;
using namespace std;
bool vis[maxn][maxn];
int move[8][2]={{-1,2},{1,2},{-1,-2},{1,-2},{-2,1},{2,1},{-2,-1},{2,-1}};
int step[maxn][maxn];
struct Map
{
    int x,y;
};


int check(int x,int y)
{
    if(x>=1&&y>=1&&y<=8&&x<=8)
        return 1;
    return 0;
}

int bfs(int sx,int sy,int ex,int ey)
{
    queue<Map>q;
    Map a;
    a.x=sx;
    a.y=sy;
    q.push(a);
    vis[sx][sy]=true;
    while(!q.empty())
    {
        Map temp=q.front();
        q.pop();
        for(int i=0;i<8;i++)
        {
            int tx=temp.x+move[i][0];
            int ty=temp.y+move[i][1];
            if(check(tx,ty)&&!vis[tx][ty])
            {
               step[tx][ty]=step[temp.x][temp.y]+1;
               vis[tx][ty]=true;
               if(tx==ex&&ty==ey)
                   return step[ex][ey];
               Map tmp;
               tmp.x=tx;
               tmp.y=ty;
               q.push(tmp);
            }
        }
    }
}

int main()
{
    int sx,sy,ex,ey;
    char s,s1,c;
    while(scanf("%c%d%c%c%d",&s,&sy,&c,&s1,&ey)!=EOF)
    {
       memset(vis,false,sizeof(vis));
       memset(step,0,sizeof(step));
       sx=s-'a'+1;
       ex=s1-'a'+1;
       step[sx][sy]=0;
       if(sx==ex&&sy==ey)
          printf("To get from %c%d to %c%d takes %d knight moves.\n",s,sy,s1,ey,0);
       else
       printf("To get from %c%d to %c%d takes %d knight moves.\n",s,sy,s1,ey,bfs(sx,sy,ex,ey));
       getchar();
    }
    return 0;
}

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#define INF 0x3f3f3f3f
const int maxn=10;
using namespace std;
bool vis[maxn][maxn];
int step[maxn][maxn];
int ans,ex,ey;
int dx[8]={-1,-1,1,1,2,2,-2,-2};
int dy[8]={2,-2,2,-2,1,-1,1,-1};


void dfs(int sx,int sy,int depth)
{
   if(sx<1||sy<1||sx>8||sy>8||depth>=ans)
        return;
   if(sx==ex&&sy==ey)
   {
       ans=min(ans,depth);
       return;
   }
   if(!vis[sx][sy])
   {
       vis[sx][sy]=true;
       step[sx][sy]=depth;
   }
   else
   {
       if(step[sx][sy]<=depth)
           return;
       step[sx][sy]=depth;
   }
   for(int i=0;i<8;i++)
   {
       int tx=sx+dx[i];
       int ty=sy+dy[i];
       dfs(tx,ty,depth+1);
   }
}


int main()
{
    char s,s1,c;
    int sx,sy;
    while(scanf("%c%d%c%c%d",&s,&sy,&c,&s1,&ey)!=EOF)
    {
       memset(vis,false,sizeof(vis));
       memset(step,INF,sizeof(step));
       sx=s-'a'+1;
       ex=s1-'a'+1;
       ans=10000;
       dfs(sx,sy,0);
       printf("To get from %c%d to %c%d takes %d knight moves.\n",s,sy,s1,ey,ans);
       getchar();
    }
    return 0;
}

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poj2243

标签：des   c   style   class   blog   code   

原文地址：http://blog.csdn.net/u014303647/article/details/27710749