HDU 5113 Black And White(搜索+剪枝)

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题意:给你一个不超过5X5的矩阵,有k种颜色,每种颜色有c[i]个,问可不可以把格子颜色涂满并且相邻格子的颜色不相同

思路:DFS搜索,不过会超时,留意到一个剪枝就是如果当前没有涂色的格子数目/2比最多还没涂的颜色数量小的话就可以直接返回了,因为肯定会相邻了


#include<bits\stdc++.h>using namespace std;int n,m,k;int mp[6][6];int color[50];bool check(int x,int y,int c){if (x-1>=1 && mp[x-1][y]==c)return false;if (y-1>=1 && mp[x][y-1]==c)return false;return true;}bool dfs(int x,int y){for (int i = 1;i<=k;i++)if ((n*m-(m*(x-1)+y-1)+1)/2<color[i])return false;for (int i = 1;i<=k;i++){if (color[i]>0 && check(x,y,i)){color[i]--;mp[x][y]=i;if (x==n && y==m)return true;if (y+1<=m){if (dfs(x,y+1))return true;}else{if (x+1<=n){if (dfs(x+1,1))return true;}}color[i]++;mp[x][y]=-1;}}return false;}int main(){    int T,cas=1;    scanf("%d",&T);    while(T--){memset(mp,-1,sizeof(mp));printf("Case #%d:\n",cas++);scanf("%d%d%d",&n,&m,&k);for (int i = 1;i<=k;i++)scanf("%d",&color[i]);if (dfs(1,1)){puts("YES");for (int i = 1;i<=n;i++){for (int j = 1;j<m;j++)printf("%d ",mp[i][j]);printf("%d\n",mp[i][m]);}}elseputs("NO");}}

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

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
 


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