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
题意：有一个N*M的方格板，现在要在上面的每个方格上涂颜色，有K种颜色，每种颜色分别涂c[1]次、c[2]次……c[K]次，c[1]+c[2]+……+c[K]=N*M
要求每个方格的颜色与其上下左右均不同，如果可以输出YES，并且输出其中的一种涂法，如果不行，输出NO；
思路：暴力搜索，但是这样会超时，可以在搜索中加入剪枝：对于剩余的方格数res，以及当前剩余的颜色可涂数必须满足（res+1）/2>=c[i]
否则在当前情况下继续向下搜得不到正确涂法；
代码如下：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int N,K,M;
int c[30];
int mp[10][10];
int check(int x,int y,int k)
{
int f=1;
if(mp[x-1][y]==k) f=0;
if(mp[x][y-1]==k) f=0;
return f;
}
int cal(int x,int y)
{
if(x>N) return 1;
int res=(N-x)*M+M-y+2; ///剩余方格数+1 ；
for(int i=1;i<=K;i++) if(res/2<c[i]) return 0; ///剪枝，某种颜色剩余方格数>(剩余方格数+1)/2 肯定不对；
for(int i=1;i<=K;i++)
{
int f=0;
if(c[i]&&check(x,y,i)){
mp[x][y]=i; c[i]--;
if(y==M) f=cal(x+1,1);
else f=cal(x,y+1);
c[i]++;
}
if(f) return 1;
}
return 0;
}
int main()
{
int T,Case=1;
cin>>T;
while(T--)
{
scanf("%d%d%d",&N,&M,&K);
for(int i=1;i<=K;i++) scanf("%d",&c[i]);
printf("Case #%d:\n",Case++);
if(!cal(1,1)) { puts("NO"); continue; }
puts("YES");
for(int i=1;i<=N;i++)
for(int j=1;j<=M;j++)
printf("%d%c",mp[i][j],(j==M)?‘\n‘:‘ ‘);
}
return 0;
}