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 c₁ + c₂ + · · · + 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

41 5 24 13 3 41 2 2 42 3 32 2 23 2 32 2 2

 

Sample Output

Case #1:NOCase #2:YES4 3 42 1 24 3 4Case #3:YES1 2 32 3 1Case #4:YES1 22 33 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;}