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

时间：2020-11-28 22:34:47

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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 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ci次，∑ci=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

 

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