Hdu 5113 Black And White, Black And White dyeing, tips, hdu5113

Hdu 5113 Black And White, Black And White dyeing, tips, hdu5113
Black And WhiteTime Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission (s): 485 Accepted Submission (s): 131
Special Judge

Problem DescriptionIn 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 shocould be used in exactly ci 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 x M ).

The second line contains K integers ci (ci> 0), denoting the number of cells where the I-th color shoshould be used.

It's guaranteed that c1 + c2 + · + cK = 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

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

Technical solution:

First mark the board in black and white, and then divide all the color types into three categories: less than or equal to (n * m + 1)/2. A1> a2> a3. If the Division is not possible, there is no solution.

Then fill in the black-marked lattice from top down in the a1 Class, and fill in the white-marked lattice from bottom up In the a2 class, fill in the remaining grids with the a3 class (the a3 class is not strictly proved to be adjacent ).

#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;}