[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 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:

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。

 1 //gaoshenbaoyou  ------ pass system test

 2 #include <iostream>

 3 #include <sstream>

 4 #include <ios>

 5 #include <iomanip>

 6 #include <functional>

 7 #include <algorithm>

 8 #include <vector>

 9 #include <string>

10 #include <list>

11 #include <queue>

12 #include <deque>

13 #include <stack>

14 #include <set>

15 #include <map>

16 #include <cstdio>

17 #include <cstdlib>

18 #include <cmath>

19 #include <cstring>

20 #include <climits>

21 #include <cctype>

22 using namespace std;

23 #define XINF INT_MAX

24 #define INF 0x3FFFFFFF

25 #define MP(X,Y) make_pair(X,Y)

26 #define PB(X) push_back(X)

27 #define REP(X,N) for(int X=0;X<N;X++)

28 #define REP2(X,L,R) for(int X=L;X<=R;X++)

29 #define DEP(X,R,L) for(int X=R;X>=L;X--)

30 #define CLR(A,X) memset(A,X,sizeof(A))

31 #define IT iterator

32 typedef long long ll;

33 typedef pair<int,int> PII;

34 typedef vector<PII> VII;

35 typedef vector<int> VI;

36 const int maxn=30;

37 int a[maxn];

38 bool flag=0;

39 int ans[10][10];

40 int n,m,k;

41 void dfs(int x,int y,int left)

42 {

43     if(!left)

44     {

45         flag=1;

46         return;

47     }

48     for(int i=1;i<=k;i++)

49         if(a[i]>(left+1)/2)return;

50     for(int i=1;i<=k;i++)

51     {

52         if(!a[i])continue;

53         if(x&&ans[x-1][y]==i)continue;

54         if(y&&ans[x][y-1]==i)continue;

55         a[i]--;

56         ans[x][y]=i;

57         if(y<m-1)dfs(x,y+1,left-1);

58         else dfs(x+1,0,left-1);

59         if(flag)return;

60         a[i]++;    

61     }

62     return;

63 }

64 int main()

65 {

66     //ios::sync_with_stdio(false);

67     int t;

68     scanf("%d",&t);

69     int cas=1;

70     while(t--)

71     {

72         flag=0;

73         scanf("%d%d%d",&n,&m,&k);

74         int sum=0;

75         int maxx=0;

76         int tot=n*m;

77         for(int i=1;i<=k;i++)

78             scanf("%d",&a[i]);

79         printf("Case #%d:\n",cas++);

80         dfs(0,0,tot);

81         if(flag)

82         {

83             printf("YES\n");

84             for(int i=0;i<n;i++)

85             {

86                 for(int j=0;j<m;j++)

87                 {

88                     if(j)printf(" ");

89                     printf("%d",ans[i][j]);

90                 }

91                 printf("\n");

92             }

93         }

94         else 

95             printf("NO\n");

96     }

97     return 0;

98 }

代码君