D - Grid Coloring（思维）

题目链接：https://atcoder.jp/contests/arc080/tasks/arc080_b

Problem Statement

We have a grid with HH rows and WW columns of squares. Snuke is painting these squares in colors 11, 22, ......, NN. Here, the following conditions should be satisfied:

• For each ii (1≤i≤N1≤i≤N), there are exactly aiai squares painted in Color ii. Here, a1+a2+...+aN=HWa1+a2+...+aN=HW.
• For each ii (1≤i≤N1≤i≤N), the squares painted in Color ii are 4-connected. That is, every square painted in Color ii can be reached from every square painted in Color ii by repeatedly traveling to a horizontally or vertically adjacent square painted in Color ii.

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≤1001≤H,W≤100
• 1≤N≤HW1≤N≤HW
• ai≥1ai≥1
• a1+a2+...+aN=HWa1+a2+...+aN=HW

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Input

Input is given from Standard Input in the following format:

HH WW
NN
a1a1 a2a2 ...... aNaN

Output

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

c11c11 ...... c1Wc1W
::
cH1cH1 ...... cHWcHW

Here, cijcij is the color of the square at the ii-th row from the top and jj-th column from the left.

---

Sample Input 1 Copy

Copy

2 2
3
2 1 1

Sample Output 1 Copy

Copy

1 1
2 3

Below is an example of an invalid solution:

1 2
3 1

This is because the squares painted in Color 11 are not 4-connected.

---

Sample Input 2 Copy

Copy

3 5
5
1 2 3 4 5

Sample Output 2 Copy

Copy

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

---

Sample Input 3 Copy

Copy

1 1
1
1

Sample Output 3 Copy

Copy

1

题意：
给你一个h*w的网格。给你n中颜色，每种颜色ai个保证所有颜色的和正好=h*w。保证每种颜色都要相连（上，下，左，右）。

找到一种方案：

代码：
 

#include
using namespace std;
int a[10009];
int b[200][200];
int main()
{
    int h,w;
    int n;
    cin>>h>>w>>n;
    for(int i=1; i<=n; i++)
    {
        cin>>a[i];
    }
    int k=1;
    for(int i=1; i<=h; i++)
    {
        if(i%2==1)
        {
            for(int j=1; j<=w; j++)
            {
                
                b[i][j]=k;
                a[k]-=1;
                if(a[k]==0)
                {
                    k+=1;
                }
            }
        }
        else
        {
            for(int j=w; j>=1; j--)
            {
               
                b[i][j]=k;
                a[k]-=1;
                if(a[k]==0)
                {
                    k+=1;
                }
            }

        }
    }
    for(int i=1; i<=h; i++)
    {
        for(int j=1; j<=w; j++)
        {
            cout<