HDU5113 Black And White
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Black And White
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 4198 Accepted Submission(s): 1148
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
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
题目大意是说给你一个N*M的矩阵,再给你K个整数填进这个矩阵,输入Ci(1<=i<=K),Ci表示第i个数字的个数。
要剪枝
#include<cstring>#include<cstdio>int n,m,k;int array[26];int map[30][30];int xnext,ynext;int count=1;int flag;void dfs(int x,int y,int z){if(z==n*m){flag=1;printf("Case #%d:\n",count++);printf("YES\n");for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(j==m){printf("%d\n",map[i][j]);break;}printf("%d ",map[i][j]);}}return;}else{for(int i=1;i<=k;i++){if((n*m-z+1)/2<array[i])//剪枝return;}for(int i=1;i<=k;i++){if(array[i]!=0&&map[x-1][y]!=i&&map[x][y-1]!=i){array[i]--;map[x][y]=i;if(y+1>m){xnext=x+1;ynext=1;}else{xnext=x;ynext=y+1;}dfs(xnext,ynext,z+1);if(flag) return;map[x][y]=0;array[i]++;}}}}int main(){int t;scanf("%d",&t);int ans=0;while(t--){scanf("%d%d%d",&n,&m,&k);flag=0;for(int i=1;i<=k;i++)scanf("%d",&array[i]);memset(map,0,4);dfs(1,1,0);if(flag==0){printf("Case #%d:\n",count++);printf("NO\n");}}}
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