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.
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 .
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.
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
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 <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define rep(i,a,n) for(int i=a;i<n;++i)
#define per(i,a,n) for(int i=n-1;i>=a;--i)
#define fi first
#define se second
#define pb push_back
#define np next_permutation
#define mp make_pair
using namespace std;
const int maxn=1e5+5;
const int maxm=1e5+5;
int t,n,m,k,flag;
//dfs 找方案剪枝
int dp[10][10],c[100];
bool check1(int x,int y,int i)
{
if(x-1>=1 && dp[x-1][y]==i) return false;
if(y-1>=1 && dp[x][y-1]==i) return false;
return true;
}
void dfs(int x,int y,int le)
{
rep(i,1,k+1) if(c[i]>(le+1)/2) return;
if(flag) return;
if(x==n+1) { flag=1; return; }
for(int i=1;i<=k;i++)
{
if(c[i])
{
if(check1(x,y,i))
{
c[i]--;
dp[x][y] = i;
if(y==m) dfs(x+1,1,le-1);
else dfs(x,y+1,le-1);
if(flag) return;
c[i]++; dp[x][y] = 0;
}
}
}
}
int main()
{
scanf("%d",&t);
rep(tt,1,t+1)
{
flag=0;
memset(dp,0,sizeof(dp));
scanf("%d%d%d",&n,&m,&k);
rep(i,1,k+1) scanf("%d",c+i);
dfs(1,1,n*m);
printf("Case #%d:\n",tt);
if(flag)
{
printf("YES");
rep(i,1,n+1)
{
rep(j,1,m) printf("%d ",dp[i][j]);
printf("%d\n",dp[i][m]);
}
}
else printf("YES");
}
return 0;
}