HDU - 5113 - Black And White（dfs）

Black And White

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

Total Submission(s): 3031    Accepted Submission(s): 831
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的格子，给出K种颜色以及每种颜色的数目，保证数目和相加 = N*M，求一种方案使得上下左右相邻的格子颜色都不相同。

思路：将颜色数目从大到小排序，记录每种颜色的id，找到一种解就返回。

wa了无数发，怀疑人生一中午，然后发现没有记录颜色id  orzzzzzzz  简直.............

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <iostream>
#include <assert.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
#define PI acos(-1.0)
const int M = 4e2 + 10;

int n, m, k;
int ans[8][8];
struct node
{
int id;
int cnt;
}color[30];

bool cmp(node a, node b)
{
return a.cnt < b.cnt;
}

bool dfs(int x, int y)
{
if (x == n + 1) return true;
if (y == m + 1) return dfs(x + 1, 1);
for (int i = k; i >= 1; i--) {
if (!color[i].cnt) continue;
if (ans[x-1][y] == color[i].id || ans[x][y-1] == color[i].id) continue;
color[i].cnt--;
ans[x][y] = color[i].id;
if (dfs(x, y + 1)) return true;
color[i].cnt++;
}
return false;
}

int main()
{
int T;
cin >> T;
for (int tt = 1; tt <= T; tt++) {
memset(ans, 0, sizeof(ans));
memset(color, 0, sizeof(color));
scanf("%d%d%d", &n, &m, &k);
for (int i = 1; i <= k; i++) {
scanf("%d", &color[i].cnt);
color[i].id = i;
}
sort(color + 1, color + 1 + k, cmp);
printf("Case #%d:\n", tt);
if(color[k].cnt > (n * m + 1) / 2){
printf("NO\n");
continue;
}
if (dfs(1, 1)) {
printf("YES\n");
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (j == m) printf("%d\n", ans[i][j]);
else printf("%d ", ans[i][j]);
}
}
}
else printf("NO\n");
}
return 0;
}