HDU 5113 Black And White (DFS + pruning)

HDU 5113 Black And White (DFS + pruning)

 

 

Question:

 

Black And White Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission (s): 1336 Accepted Submission (s): 350
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 shocould 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 x M ).

The second line contains K integers ci (ci> 0), denoting the number of cells where the I-th color shoshould 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

41 5 24 13 3 41 2 2 42 3 32 2 23 2 32 2 2

Sample Output

Case #1:NOCase #2:YES4 3 42 1 24 3 4Case #3:YES1 2 32 3 1Case #4:YES1 22 33 1

Source 2014ACM/ICPC Asia Beijing site

Solution:
Because n and m are relatively small, they will think of the search, but the query times out. You need to add a pruning because when the remaining capacity is n, any color can only be (n + 1)/2. If the number of colors exceeds this value, return. This optimization is already amazing. I added a useless optimization. The main function shows that two colors are the same, so you do not need to calculate them again.

Code:

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include #include 
        
         #include 
         
          #define LL long longusing namespace std;int map [7][7],color[30];int n,m,t,k,sz,amount,tmp;bool flag=false;void init(){for(int i=0;i<7;i++)map[0][i]=-1;for(int j=1;j<7;j++)map[j][0]=-1;}void dfs(int x,int y,int typ,int lef){ tmp=(lef+1)/2; for(int i=1;i<=k;i++) { if(color[i]>tmp) return; } if(flag)return; if(color[typ]==0)return; if(map[x-1][y]!=typ&&map[x][y-1]!=typ) { map[x][y]=typ; color[typ]--; if(x==n&&y==m) { flag=true; return; } else if(y==m) { tmp=0; for(int i=1;i<=k;i++) { if(color[i])tmp++; } if(tmp==1) { if(m!=1) { color[typ]++; return; } } for(int i=1;i<=k;i++) { if(color[i]) dfs(x+1,1,i,lef-1); } } else { for(int i=1;i<=k;i++) { if(color[i]) { dfs(x,y+1,i,lef-1); } } } color[typ]++; } return;}int main(){init();scanf(%d,&t);for(int i=1;i<=t;i++){ set 
          
            cnt; printf(Case #%d:,i); scanf(%d%d%d,&n,&m,&k); amount=(n*m+1)/2; flag=true; if(k==1) { if(n*m!=1) flag=false; } for(int j=1;j<=k;j++) { scanf(%d,&color[j]); if(color[j]>amount) flag=false; } if(flag) { flag=false; for(int j=1;j<=k;j++) { sz=cnt.size(); if(color[j]) { cnt.insert(color[j]); if(cnt.size()>sz) dfs(1,1,j,n*m); } } } if(flag) { printf(YES); for(int j=1;j<=n;j++) { printf(%d,map[j][1]);for(int k=2;k<=m;k++)printf( %d,map[j][k]);printf(); } } else { printf(NO); }}return 0;}