Black And White HDU - 5113

In mathematics, the four color theorem, or the four color map theorem, states that, given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colors are required to color the regions of the map so that no two adjacent regions have the same color. 
— Wikipedia, the free encyclopedia 

In this problem, you have to solve the 4-color problem. Hey, I’m just joking. 

You are asked to solve a similar problem: 

Color an N × M chessboard with K colors numbered from 1 to K such that no two adjacent cells have the same color (two cells are adjacent if they share an edge). The i-th color should be used in exactly c _i cells. 

Matt hopes you can tell him a possible coloring.

InputThe first line contains only one integer T (1 ≤ T ≤ 5000), which indicates the number of test cases. 

For each test case, the first line contains three integers: N, M, K (0 < N, M ≤ 5, 0 < K ≤ N × M ). 

The second line contains K integers c _i (c _i > 0), denoting the number of cells where the i-th color should be used. 

It’s guaranteed that c ₁ + c ₂ + · · · + c _K = N × M . 
OutputFor each test case, the first line contains “Case #x:”, where x is the case number (starting from 1). 

In the second line, output “NO” if there is no coloring satisfying the requirements. Otherwise, output “YES” in one line. Each of the following N lines contains M numbers seperated by single whitespace, denoting the color of the cells. 

If there are multiple solutions, output any of them.Sample Input

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

Sample Output

Case #1:
NO
Case #2:
YES
4 3 4
2 1 2
4 3 4
Case #3:
YES
1 2 3
2 3 1
Case #4:
YES
1 2
2 3
3 1
DFS

#include <iostream>
using namespace std;
#include<string.h>
#include<set>
#include<stdio.h>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<queue>
int a[30][30],yanse[30];
int lenx,leny,yanseshu;
int flag=0;
int b[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int panduan(int x,int y,int k)
{
    if(a[x-1][y]==k) return 0;
    if(a[x][y-1]==k) return 0;
    return 1;
}
int dfs(int x,int y)
{
    if(x>leny)
    return 1;
    int shengyu=(leny-x)*lenx+lenx-y+2;
    for(int i=1;i<=yanseshu;i++)
    if(shengyu/2<yanse[i])
    return 0;
    for(int i=1;i<=yanseshu;i++)
    {
        int f=0;
        if(yanse[i]&&panduan(x,y,i)){
            a[x][y]=i;
            yanse[i]--;
            if(y==lenx)
            f=dfs(x+1,1);
            else
             f=dfs(x,y+1);
            yanse[i]++;
        }
        if(f)
            return 1;
    }
    return 0;
}
int main()
{
    int add=0,t;
    cin>>t;
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(yanse,0,sizeof(yanse));
        cin>>leny>>lenx>>yanseshu;
        for(int i=1;i<=yanseshu;i++)
            cin>>yanse[i];
            cout<<"Case #"<<++add<<":"<<endl;
            flag=0;
            if(dfs(1,1))
            {
                cout<<"YES"<<endl;
                for(int i=1;i<=leny;i++)
                {
                    for(int j=1;j<=lenx;j++)
                    {
                        if(j==lenx)
                        {
                            cout<<a[i][j];
                            continue;
                        }
                        cout<<a[i][j]<<' ';
                    }
                    cout<<endl;
                }
            }
            else
                    cout<<"NO"<<endl;
    }
    return 0;
}