「题解」：砖块

Posted 2022-12-28 xingmi-weiyouni

问题 A: 砖块

时间限制: 1 Sec  内存限制: 256 MB

题面

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题面谢绝公开。

题解

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大模拟需要题解么？？

 

#include<bits/stdc++.h>
#define rint register int
using namespace std;
int T,len,vis[1111][1111],ans;
int l;//l表示长边在哪个方向，1表南北，2表东西，0表底面在下。
int x,y;//左下角在网格中占据的坐标
int h;//高度
char ch[103];
inline int Max(int A,int B)return A>B?A:B;
inline void North()if(l==0)y+=1;l=1;else if(l==1)y+=h;l=0;else y+=1;
inline void South()if(l==0)y-=h;l=1;else if(l==1)y-=1;l=0;else y-=1;
inline void East()if(l==0)x+=1;l=2;else if(l==1)x+=1;elsex+=h;l=0;
inline void West()if(l==0)x-=h;l=2;else if(l==1)x-=1;elsex-=1;l=0;
int main()

//    freopen("ex_block2.in","r",stdin);
//    freopen("my.out","w",stdout);
    scanf("%d",&T);
    while(T--)
    
        l=x=y=len=ans=0;
        memset(vis,0,sizeof(vis));
        scanf("%d %s",&h,ch+1);
        len=strlen(ch+1);vis[555][555]=1;
        for(rint i=1;i<=len;++i)
        
            switch(ch[i])
            
                case ‘N‘:North();break;
                case ‘S‘:South();break;
                case ‘E‘:East(); break;
                case ‘W‘:West(); break;
            
            if(l==0)
            
                vis[x+555][y+555]++;
                ans=Max(ans,vis[x+555][y+555]);
            
            else if(l==1)
            
                for(rint i=0;i<h;++i)
                
                    vis[x+555][y+i+555]++;
                    ans=Max(ans,vis[x+555][y+i+555]);
                
            
            else
            
                for(rint i=0;i<h;++i)
                
                    vis[x+i+555][y+555]++;
                    ans=Max(ans,vis[x+i+555][y+555]);
                
            
        
        if(l==0)
            printf("%d\n%d\n%d\n",x,y,ans);
        else if(l==1)
        
            for(rint i=0;i<h;++i)printf("%d ",x);puts("");
            for(rint i=0;i<h;++i)printf("%d ",y+i);puts("");
            printf("%d\n",ans);
        
        else 
        
            for(rint i=0;i<h;++i)printf("%d ",x+i);puts("");
            for(rint i=0;i<h;++i)printf("%d ",y);puts("");
            printf("%d\n",ans);
        
    
    return 0;

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