AtCoder Grand Contest 021C-Tiling-CFANZ编程社区

C - Tiling

Time limit : 2sec / Memory limit

Score : 900

Problem Statement

Takahashi has an N×M grid, with N horizontal rows and M vertical columns. Determine if we can place A 1×2 tiles (1 vertical, 2 horizontal) and B 2×1 tiles (2vertical, 1

All the tiles must be placed on the grid.
Tiles must not stick out of the grid, and no two different tiles may intersect.
Neither the grid nor the tiles may be rotated.
Every tile completely covers exactly two squares.

Constraints

1≤N,M≤1000
0≤A,B≤500000
N, M, A and B

Input

Input is given from Standard Input in the following format:

N M A B

Output

If it is impossible to place all the tiles, print NO. Otherwise, print the following:

YES c11…c1M : cN1…cNM

Here, cij must be one of the following characters: ., <, >, ^ and v. Represent an arrangement by using each of these characters as follows:

When cij is

.

, it indicates that the square at the i-th row and j-th column is empty;
When cij is

<

, it indicates that the square at the i-th row and j-th column is covered by the left half of a 1×2
When cij is

>

, it indicates that the square at the i-th row and j-th column is covered by the right half of a 1×2
When cij is

^

, it indicates that the square at the i-th row and j-th column is covered by the top half of a 2×1
When cij is

v

, it indicates that the square at the i-th row and j-th column is covered by the bottom half of a 2×1

Sample Input 1

Copy

3 4 4 2

Sample Output 1

Copy

YES <><> ^<>^ v<>v

This is one example of a way to place four 1×2 tiles and three 2×1 tiles on a 3×4

Sample Input 2

Copy

4 5 5 3

Sample Output 2

Copy

YES <>..^ ^.<>v v<>.^ <><>v

Sample Input 3

Copy

7 9 20 20

Sample Output 3

Copy

Submit

模拟题坑死人！！！

题解

Code：

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#define N 1005
using namespace std;
char ans[N][N];
inline int read()  
{  
    int x=0,f=1;char s=getchar();  
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}  
    while(s<='9'&&s>='0'){x=x*10+s-'0';s=getchar();}  
    return x*f;  
}  
int main() 
{ 
    int n=read(),m=read(),a=read(),b=read();
    memset(ans,'.',sizeof(ans));
    for(int i=1;i<=n;i++)ans[i][m+1]='\0';
    if(n&1) 
    {
        int j=1;
        while(a&&j+1<=m) 
        {
            ans[n][j]='<';
            ans[n][j+1]='>';
            j+=2;
            a--;
        }
    }
    if(m&1) 
    {
        int i=n;
        while(b&&i>1)
        {
            ans[i-1][m]='^';
            ans[i][m]='v';
            i-=2;
            b--;
        }
    }
    for(int i=n/2;i>=1;i--)
        for(int j=1;j<=m/2;j++) 
        {
            int x=i*2-1,y=j*2-1;
            if(n%2==1&&m%2==1&&a==1&&b==1&&i==1&&j==m/2) 
            {
                ans[x][y]='^';
                ans[x+1][y]='v';
                b--;
                ans[x][y+1]='<';
                ans[x][y+2]='>';
                a--;
            }
            if(a&&b<2) 
            {
                ans[x][y]='<';
                ans[x][y+1]='>';
                a--;
                if(a)
                {
                    ans[x+1][y]='<';
                    ans[x+1][y+1]='>';
                    a--;
                }
            }else if(b) 
            {
                ans[x][y]='^';
                ans[x+1][y]='v';
                b--;
                if(b)
                {
                    ans[x][y+1]='^';
                    ans[x+1][y+1]='v';
                    b--;
                }
            }
        }
    if(a||b)puts("NO");else
    {
        puts("YES");
        for(int i=1;i<=n;i++)puts(ans[i]+1);
    }
    return 0;
}