HDU 5113 Black And White
Black And White
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 5859 Accepted Submission(s): 1615
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.
— 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 .
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.
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
Source
Recommend
liuyiding
题解:
搜索+减枝优化(判断是否有颜色的数量超过剩下的方格数量的一半)
参考代码:
1 #include <bits/stdc++.h> 2 #define INF 0x3f3f3f3f 3 #define eps 1e-9 4 #define rep(i,a,n) for(int i=a;i<n;++i) 5 #define per(i,a,n) for(int i=n-1;i>=a;--i) 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define np next_permutation 10 #define mp make_pair 11 using namespace std; 12 const int maxn=1e5+5; 13 const int maxm=1e5+5; 14 15 int t,n,m,k,flag; 16 //dfs 找方案剪枝 17 int dp[10][10],c[100]; 18 bool check1(int x,int y,int i) 19 { 20 if(x-1>=1 && dp[x-1][y]==i) return false; 21 if(y-1>=1 && dp[x][y-1]==i) return false; 22 return true; 23 } 24 void dfs(int x,int y,int le) 25 { 26 rep(i,1,k+1) if(c[i]>(le+1)/2) return; 27 if(flag) return; 28 if(x==n+1) { flag=1; return; } 29 for(int i=1;i<=k;i++) 30 { 31 if(c[i]) 32 { 33 if(check1(x,y,i)) 34 { 35 c[i]--; 36 dp[x][y] = i; 37 if(y==m) dfs(x+1,1,le-1); 38 else dfs(x,y+1,le-1); 39 if(flag) return; 40 c[i]++; dp[x][y] = 0; 41 } 42 } 43 } 44 } 45 int main() 46 { 47 scanf("%d",&t); 48 rep(tt,1,t+1) 49 { 50 flag=0; 51 memset(dp,0,sizeof(dp)); 52 scanf("%d%d%d",&n,&m,&k); 53 rep(i,1,k+1) scanf("%d",c+i); 54 dfs(1,1,n*m); 55 printf("Case #%d:\n",tt); 56 if(flag) 57 { 58 printf("YES"); 59 rep(i,1,n+1) 60 { 61 rep(j,1,m) printf("%d ",dp[i][j]); 62 printf("%d\n",dp[i][m]); 63 } 64 } 65 else printf("YES"); 66 } 67 return 0; 68 } 69