K

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

  解法:

#include <iostream>
#include<string.h>
#include<stdio.h>

using namespace std;

const int MAX = 30 +19;
int Map[MAX][MAX],a[MAX];
int L,W,n;
int pand(int x,int y,int i)
{
    if(Map[x-1][y]==i) return 0;
    if(Map[x][y-1]==i) return 0;
    return 1;
}
int DFS(int x,int y)
{
    if(x>L)
    return 1;

    for(int i=1;i<=n;i++)
    if(((L-x)*W+W-y+2)/2<a[i])
    return 0;

    for(int i=1;i<=n;i++)
    {
        int TX=0;
        if(a[i]&&pand(x,y,i)){
            Map[x][y]=i;
            a[i]--;
            if(y==W)
            TX = DFS(x+1,1);
            else
             TX = DFS(x,y+1);
            a[i]++;
        }
        if(TX)
            return 1;
    }
    return 0;
}
int main()
{
    int N0=0,N;
    cin>>N;
    while(N--)
    {
        memset(a,0,sizeof(a));
        memset(Map,0,sizeof(Map));

        cin>>L>>W>>n;
        for(int i=1;i<=n;i++)
            cin>>a[i];
            cout<<"Case #"<<++N0<<":"<<endl;

            if(DFS(1,1))
            {
                cout<<"YES"<<endl;
                for(int i=1;i<=L;i++)
                {
                    cout<<Map[i][1];
                    for(int j=2;j<=W;j++)
                    {
                        cout<<" "<<Map[i][j];
                    }
                    cout<<endl;
                }
            }
            else
                    cout<<"NO"<<endl;
    }
    return 0;
}