Grid Coloring(AtCoder-2687)

Problem Description

We have a grid with H rows and W columns of squares. Snuke is painting these squares in colors 1, 2, …, N. Here, the following conditions should be satisfied:

For each i (1≤i≤N), there are exactly ai squares painted in Color i. Here, a1+a2+…+aN=HW.
For each i (1≤i≤N), the squares painted in Color i are 4-connected. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i.
Find a way to paint the squares so that the conditions are satisfied. It can be shown that a solution always exists.

Constraints

  • 1≤H,W≤100
  • 1≤NHW
  • ai≥1
  • a1+a2+…+aN=HW

Input

Input is given from Standard Input in the following format:

H W
N
a1 a2 … aN

Output

Print one way to paint the squares that satisfies the conditions. Output in the following format:

c11 … c1W
:
cH1 … cHW
Here, cij is the color of the square at the i-th row from the top and j-th column from the left.

Example

Sample Input 1

2 2
3
2 1 1

Sample Output 1

1 1
2 3
Below is an example of an invalid solution:

1 2
3 1
This is because the squares painted in Color 1 are not 4-connected.

Sample Input 2

3 5
5
1 2 3 4 5

Sample Output 2

1 4 4 4 3
2 5 4 5 3
2 5 5 5 3

Sample Input 3

1 1
1
1

Sample Output 3

1

题意:有一个 h*w 的区域,要在这些方格上涂 n 种颜色,第 i 种颜色涂 ai 个,对于每一种颜色,其涂得方块均可以通过上下左右移动到达同种颜色的区域,即相同颜色涂得方块是连通的,输出一种涂色方法

思路:由于要求同种颜色是连通的,且任意输出一种方法,那么从对于 h*w 个方格,从第一个开始从上到下走 S 型涂色即可

Source Program

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 100+5;
const int dx[] = {-1,1,0,0,-1,-1,1,1};
const int dy[] = {0,0,-1,1,-1,1,-1,1};
using namespace std;

int a[N*N];
int G[N][N];
int main() {
    int h,w,n;
    scanf("%d%d",&h,&w);
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);

    int cnt=1;
    for(int i=1; i<=h; i++) {
        if(i%2) {
            for(int j=1; j<=w; j++) {
                if(a[cnt]==0) {
                    cnt++;
                    G[i][j]=cnt;
                    a[cnt]--;
                }
                else if(a[cnt]>0) {
                    G[i][j]=cnt;
                    a[cnt]--;
                }
            }
        }
        else {
            for(int j=w; j>=1; j--) {
                if(a[cnt]==0) {
                    cnt++;
                    G[i][j]=cnt;
                    a[cnt]--;
                }
                else if(a[cnt]>0) {
                    G[i][j]=cnt;
                    a[cnt]--;
                }
            }
        }
    }

    for(int i=1; i<=h; i++) {
        for(int j=1; j<=w; j++)
            printf("%d ",G[i][j]);
        printf("\n");
    }
    return 0;
}

 

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