hdu 5113 Black And White

Black And White

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Total Submission(s): 1173    Accepted Submission(s): 306

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

 
题意：给一个n*m的矩阵和k种染料，每种染料要染ai个方格，保证a1 + a2 +...+ ak = n * m，给这个矩阵染色，要求相邻方格颜色不能相同。
思路：没想到是个简单的剪枝搜索。。。只需要在开始判断一下颜色最多的方格个数是不是超过了（n + m + 1）/ 2就可以了。。。

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct Color {
int color, n;
bool operator < (const Color &i) const {
return n > i.n;
}
}color[30];
int mp[10][10], n, m, k;
bool dfs(int x, int y) {
if(x == n) return true;
if(y == m) return dfs(x + 1, 0);
for(int i = 1; i <= k; i++) {
if(color[i].n) {
if(x && mp[x - 1][y] == color[i].color) continue;
if(y && mp[x][y - 1] == color[i].color) continue;
mp[x][y] = color[i].color;
color[i].n--;
if(dfs(x, y + 1)) return true;
color[i].n++;
}
}
return false;
}
main() {
int t;
scanf("%d", &t);
for(int cas = 1; cas <= t; cas++) {
scanf("%d %d %d",&n, &m, &k);
for(int i = 1; i <= k; i++) {
scanf("%d", &color[i].n);
color[i].color = i;
}
sort(color + 1, color + 1 + k);
printf("Case #%d:\n", cas);
if(color[1].n > (n * m + 1) / 2) {
puts("NO");
continue;
}
if(dfs(0,0)) {
puts("YES");
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++)
if(!j) printf("%d", mp[i][j]);
else printf(" %d", mp[i][j]);
putchar('\n');
}
}
else puts("NO");
}
}