# 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 ci 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 ci (ci > 0), denoting the number of cells where the i-th color should be used.

It’s guaranteed that c1 + c2 + · · · + cK = 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
```

```#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;
}
```

## HDU 1565 （最大流+黑白染色化二分图求最小割）

http://acm.hdu.edu.cn/showproblem.php?pid=1565 思路:将横纵坐标和为偶尔染白色,其他染黑色,黑点连接源点,流量为该点的值,白点连接汇点,流量为该点的值,黑白点有相邻的就连边,值为无穷大.最后求最大流,即该图的最小割. PS:刚开始不明白为为什么最大流会等于最小割,为什么所有的点之和减去最小割就会等于答案. 我的理解是:整张图其实就跟连接管道一样,连接了黑点表示取了黑点那个值的流量,白点也是,而连接了相邻的黑白点求出的最大流就会是流量较小的那个的值.好

## hdu 5113 Black And White

http://acm.hdu.edu.cn/showproblem.php?pid=5113 题意:给你n*m的格子,然后在每个格子内涂色,相邻格子不能同色,然后给你每个颜色涂的格子的固定个数,然后可不可以实现,可以实现输出任意一种,否则输出NO 思路:dfs枚举,剪纸,每种颜色剩余的个数不能超过剩余格子数的一半,如果剩余格子数是奇数,不能超过一半加1,偶数是一半. 1 #include <cstdio> 2 #include <cstring> 3 #include <al

## hdu 5113 Black And White （dfs回溯+剪枝）

Black And White Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 854    Accepted Submission(s): 218 Special Judge Problem Description In mathematics, the four color theorem, or the four color

## 搜索（剪枝优化）：HDU 5113 Black And White

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 th