HDUoj 5113 Black And White （dfs&剪枝

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

It’s guaranteed that c 1 + c 2 + · · · + c K = 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:
4000

YES
1 2 3
2 3 1
Case #4:
YES
1 2
2 3
3 1

题意

给一个格子和几种颜色与数量 要求相邻点颜色不能相同

题解：

剪枝方法 就是剩余格子数/2必定大于max(c[i])

AC代码

#include <bits/stdc++.h>
using namespace std;

#define LL long long
#define CLR(a,b) memset(a,(b),sizeof(a))

int mps[10][10];
int c[100];
bool flag;
int n, m, k;
void dfs(int x,int y,int color)
{
int res = (n-x)*m+m-y+1;
for(int i = 1; i <= k; i++) if(c[i] > res/2) return;   //剪枝
mps[x][y] = color;
if(x==n && y==m) {
flag = true; return;
}
if(y+1 <= m) {
int dx =x;
int dy = y+1;
for(int i = 1; i<= k; i++) {
if(c[i] && i!=mps[dx][dy-1] && i!=mps[dx-1][dy]) {
c[i]--;
dfs(dx,dy,i);
if(flag) return;
c[i]++;
}
}
}
else if(x+1 <= n) {
int dx = x+1;
int dy = 1;
for(int i = 1; i <= k; i++) {
if(c[i] && i!=mps[dx-1][dy]) {
c[i]--;
dfs(dx,dy,i);
if(flag) return;
c[i]++;
}
}
}
}
int main()
{
int zz = 0;
int T;
scanf("%d",&T);
while(T--) {
scanf("%d%d%d",&n,&m,&k);
for(int i = 1; i <= k; i++) {
scanf("%d",&c[i]);
}
flag = false;
CLR(mps,0);
for(int i = 1; i <= k; i++) {
c[i]--;
dfs(1,1,i);
if(flag) break;
c[i]++;
}
if(flag) {
printf("Case #%d:\nYES\n", ++zz);
for(int i = 1; i<= n; i++) {
for(int j = 1; j <= m; j++) {
if(j == m)
printf("%d\n",mps[i][j]);
else
printf("%d ",mps[i][j]);
}
}
}
else {
printf("Case #%d:\nNO\n", ++zz);
}
}
return 0;
}