HDU-5113 Black And White

                        Black And White

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

 同14年北京站

题意：t组数据，每组n,m,k. n行m列，k种颜色（1~k），然后输出一行1~k种颜色所能使用的次数。

相同颜色不能相邻，上下左右。 如果可以输出YES和摆放的方法，否则NO

思路：DFS+剪枝，从左上往右下搜索，按Z字型来搜

剪枝:当剩下的颜色数大于所剩格子的一半，那么就直接return了。

开始一直想的是搜四个方向，后面卡了很久，想到只要从左下搜到右下就行了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6+10;
const int MOD=1e9+7;
int n,m,k;
int c[30];
int mapp[6][6];
bool flag;
bool check(int x,int y,int color)
{
    if(c[color]&&mapp[x-1][y]!=color&&mapp[x][y-1]!=color)
        return true;
    return false;
}
void DFS(int x,int y,int step)
{
    for(int i=1; i<=k; i++)
    {
        if(c[i]>((step+1)/2))
            return;
    }//没剪枝T
    if(step == 0)
    {
        flag = true;
    }
    if(flag)
        return;
    for(int i = 1; i <= k; i++)
    {
        if(check(x,y,i))
        {
            mapp[x][y] = i;
            c[i]--;
            if(y+1>m)
                DFS(x + 1,1,step-1);
            else
                DFS(x,y + 1,step-1);
            if(flag)
                return;
            c[i]++;
            mapp[x][y] = -1;
        }
    }
}
int main()
{
    int T;
    int p=0;
    scanf("%d",&T);
    while(T--)
    {
        flag = false;
        memset(mapp,-1,sizeof(mapp));
        scanf("%d%d%d",&n,&m,&k);
        for(int i = 1; i <= k; i++)
            scanf("%d",&c[i]);
        DFS(1,1,n*m);
        printf("Case #%d:\n",++p);
        if(flag)
        {
            cout << "YES" << endl;
            for(int i = 1; i <= n; i++)
            {
                for(int j = 1; j <= m; j++)
                    j==m?cout << mapp[i][j] << endl:cout<<mapp[i][j]<<" ";//wa了一发
            }
        }
        else
            cout << "NO" << endl;
    }
    return 0;
}

 

 

 

PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~