hdu 5113 Black And White, 黑白染色，技巧

Black And White

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 485    Accepted Submission(s): 131
Special Judge

Problem Description

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.

 

Input

The 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 .

 

Output

For 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
   
   
   
   
   
   
   
   


   
   
   
   

技巧性解法：

先对棋盘黑白标记，然后把所有的颜色种类分成小于等于（n*m+1)/2的三类。 a1 > a2 > a3 。无法划分则无解。

然后a1类从上往下填黑色标记的格子，a2类从下往上填白色标记的格子，剩余的格子用a3类填(经过不严格证明 a3类的不会相邻)。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
const int maxn=51;
int i,j,k;
int n,m,nm,nm1,nm2,bj;
int a[maxn],wz1[maxn],ans[maxn];
int b[maxn],wz2[maxn],col[maxn];
int line[4][maxn],gs[4];
int cmp(int x,int y){
    return gs[x]>gs[y];
}
int main(){
    int T,q;
    scanf("%d",&T);
    for(int ca=1;ca<=T;++ca){
        printf("Case #%d:\n",ca);
        scanf("%d%d%d",&n,&m,&q);
        for(i=1;i<=q;++i){
            scanf("%d",&a[i]);
        }
        nm=n*m;
        nm1=(nm+1)>>1;
        nm2=nm>>1;
        
        int l1=0,l2=0;
        col[0]=0;
        for(int i=2;i<=m;++i)    col[i]=1-col[i-1];
        for(int i=m+1;i<=nm;++i) col[i]=1-col[i-m];
        for(int i=1;i<=nm;++i){
            if(col[i])wz2[++l2]=i;
            else wz1[++l1]=i;
        }
        
        
        gs[1]=gs[2]=gs[3]=0;
        for(i=1;i<4;++i)b[i]=i;
        for(i=1;i<=q && gs[1]+a[i]<=nm1;++i){
            while(a[i]--)line[1][++gs[1]]=i;
        }
        for(;i<=q && gs[2]+a[i]<=nm1;++i){
            while(a[i]--)line[2][++gs[2]]=i;
        }
        for(;i<=q && gs[3]+a[i]<=nm1;++i){
            while(a[i]--)line[3][++gs[3]]=i;
        }
        
        if(i<=q){
            puts("NO");
            continue;
        }
        puts("YES");
        sort(b+1,b+4,cmp);
        for(i=1;i<=nm1-gs[b[1]];++i){
            ans[wz1[i]]=line[b[3]][i];
        }
        k=i;
        for(j=1;j<=gs[b[1]];++j,++i){
            ans[wz1[i]]=line[b[1]][j];
        }
        for(i=nm2;i>gs[b[2]];--i,++k){
            ans[wz2[i]]=line[b[3]][k];
        }
        for(;i>0;--i){
            ans[wz2[i]]=line[b[2]][i];
        }
        for(int i=0;i<n;++i){
            for(int j=0;j<m;++j){                
                if(j)printf(" ");
                printf("%d",ans[i*m+1+j]);
            }
            puts("");
        }
    }
//    system("pause");
    return 0;
}