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题面：
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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 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
Source 2014ACM/ICPC亚洲区北京站
解题：
因为n,m比较小，会想到搜索，但直接超时了。需要加一个剪枝，因为当剩下的容量为n时，任意一种颜色最多只能为(n+1)/2。当任意一种颜色数量超过这个值时，就返回。这个优化已经很厉害了。我又加了一个没什么用的优化，main函数中出现某两种颜色相同时，不用重复计算。
代码：
#include#include #include #include #include #include #include #include #include #include #define LL long long using 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; }
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