題目鏈接 Problem Description In mathematics, the four color theorem, or the four color map theorem, states that, given any separation of a plane into cont ...
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.
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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 .
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
題意：有一個N*M的方格板，現在要在上面的每個方格上塗顏色，有K種顏色，每種顏色分別塗c[1]次、c[2]次……c[K]次，c[1]+c[2]+……+c[K]=N*M
要求每個方格的顏色與其上下左右均不同，如果可以輸出YES，並且輸出其中的一種塗法，如果不行，輸出NO；
思路：暴力搜索，但是這樣會超時，可以在搜索中加入剪枝：對於剩餘的方格數res，以及當前剩餘的顏色可塗數必須滿足（res+1）/2>=c[i]
否則在當前情況下繼續向下搜得不到正確塗法；
代碼如下：
#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; ///剩餘方格數+1 ； for(int i=1;i<=K;i++) if(res/2<c[i]) return 0; ///剪枝，某種顏色剩餘方格數>(剩餘方格數+1)/2 肯定不對； 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; }
