HDU 5113--Black And White(搜索+剪枝)
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
题意：有一个N*M的方格板，现在要在上面的每个方格上涂颜色，有K种颜色，每种颜色分别涂c[1]次、c[2]次……c[K]次，c[1]+c[2]+……+c[K]=N*M
要求每个方格的颜色与其上下左右均不同，如果可以输出YES，并且输出其中的一种涂法，如果不行，输出NO；
思路：暴力搜索，但是这样会超时，可以在搜索中加入剪枝：对于剩余的方格数res，以及当前剩余的颜色可涂数必须满足（res+1）/2>=c[i]
否则在当前情况下继续向下搜得不到正确涂法；
代码如下：
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; int N,K,M; int c[30]; int mp[10][10]; int check(int x,int y,int k) { int f=1; if(mp[x-1][y]==k) f=0; if(mp[x][y-1]==k) f=0; return f; } int cal(int x,int y) { if(x>N) return 1; int res=(N-x)*M+M-y+2; ///剩余方格数+1 ； for(int i=1;i<=K;i++) if(res/2<c[i]) return 0; ///剪枝，某种颜色剩余方格数>(剩余方格数+1)/2 肯定不对； for(int i=1;i<=K;i++) { int f=0; if(c[i]&&check(x,y,i)){ mp[x][y]=i; c[i]--; if(y==M) f=cal(x+1,1); else f=cal(x,y+1); c[i]++; } if(f) return 1; } return 0; } int main() { int T,Case=1; cin>>T; while(T--) { scanf("%d%d%d",&N,&M,&K); for(int i=1;i<=K;i++) scanf("%d",&c[i]); printf("Case #%d:\n",Case++); if(!cal(1,1)) { puts("NO"); continue; } puts("YES"); for(int i=1;i<=N;i++) for(int j=1;j<=M;j++) printf("%d%c",mp[i][j],(j==M)?'\n':' '); } return 0; }
