[hdu5113]Black And White2014北京赛区现场赛B题（搜索加剪枝）

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Black And White

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
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 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 c₁ + c₂ + · · · + 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:

YES

1 2 3

2 3 1

Case #4:

YES

1 2

2 3

3 1

 

题意：有一个n*m个地图，用k中颜色来进行填充，每种颜色可以使用的次数为c_i次，∑c_i=n*m,要求相邻的格子的颜色不能相同，问是否存在满足要求的染色方案，若存在，则输出其中一种。

分析：注意到n,m≤5,图较小，考虑用dfs来搞，但是光是dfs会T，所以需要加上一个剪枝。

若当前某种颜色的剩余数目大于剩余格子数目的一半，则必定不能完成填充方案，直接跳出。

 //gaoshenbaoyou  ------ pass system test
 #include <iostream>
 #include <sstream>
 #include <ios>
 #include <iomanip>
 #include <functional>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <list>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <set>
 #include <map>
 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <climits>
 #include <cctype>
 using namespace std;
 #define XINF INT_MAX
 #define INF 0x3FFFFFFF
 #define MP(X,Y) make_pair(X,Y)
 #define PB(X) push_back(X)
 #define REP(X,N) for(int X=0;X<N;X++)
 #define REP2(X,L,R) for(int X=L;X<=R;X++)
 #define DEP(X,R,L) for(int X=R;X>=L;X--)
 #define CLR(A,X) memset(A,X,sizeof(A))
 #define IT iterator
 typedef long long ll;
 typedef pair<int,int> PII;
 typedef vector<PII> VII;
 typedef vector<int> VI;
 const int maxn=;
 int a[maxn];
 bool flag=;
 int ans[][];
 int n,m,k;
 void dfs(int x,int y,int left)
 {
     if(!left)
     {
         flag=;
         return;
     }
     for(int i=;i<=k;i++)
         if(a[i]>(left+)/)return;
     for(int i=;i<=k;i++)
     {
         if(!a[i])continue;
         if(x&&ans[x-][y]==i)continue;
         if(y&&ans[x][y-]==i)continue;
         a[i]--;
         ans[x][y]=i;
         if(y<m-)dfs(x,y+,left-);
         else dfs(x+,,left-);
         if(flag)return;
         a[i]++;
     }
     return;
 }
 int main()
 {
     //ios::sync_with_stdio(false);
     int t;
     scanf("%d",&t);
     int cas=;
     while(t--)
     {
         flag=;
         scanf("%d%d%d",&n,&m,&k);
         int sum=;
         int maxx=;
         int tot=n*m;
         for(int i=;i<=k;i++)
             scanf("%d",&a[i]);
         printf("Case #%d:\n",cas++);
         dfs(,,tot);
         if(flag)
         {
             printf("YES\n");
             for(int i=;i<n;i++)
             {
                 for(int j=;j<m;j++)
                 {
                     if(j)printf(" ");
                     printf("%d",ans[i][j]);
                 }
                 printf("\n");
             }
         }
         else
             printf("NO\n");
     }
     return ;
 }

代码君