HDU 5113 Black And White
阿新 • • 發佈:2018-09-27
reg cal bmi ase its next free this use
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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.
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 .
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
Black And White
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 5859 Accepted Submission(s): 1615
Special Judge
— 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
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
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 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亞洲區北京站-重現賽(感謝北師和上交) Recommend liuyiding 題解: 搜索+減枝優化(判斷是否有顏色的數量超過剩下的方格數量的一半) 參考代碼:
#include <bits/stdc++.h> #define INF 0x3f3f3f3f #define eps 1e-9 #define rep(i,a,n) for(int i=a;i<n;++i) #define per(i,a,n) for(int i=n-1;i>=a;--i) #define fi first #define se second #define pb push_back #define np next_permutation #define mp make_pair using namespace std; const int maxn=1e5+5; const int maxm=1e5+5; int t,n,m,k,flag; //dfs 找方案剪枝 int dp[10][10],c[100]; bool check1(int x,int y,int i) { if(x-1>=1 && dp[x-1][y]==i) return false; if(y-1>=1 && dp[x][y-1]==i) return false; return true; } void dfs(int x,int y,int le) { rep(i,1,k+1) if(c[i]>(le+1)/2) return; if(flag) return; if(x==n+1) { flag=1; return; } for(int i=1;i<=k;i++) { if(c[i]) { if(check1(x,y,i)) { c[i]--; dp[x][y] = i; if(y==m) dfs(x+1,1,le-1); else dfs(x,y+1,le-1); if(flag) return; c[i]++; dp[x][y] = 0; } } } } int main() { scanf("%d",&t); rep(tt,1,t+1) { flag=0; memset(dp,0,sizeof(dp)); scanf("%d%d%d",&n,&m,&k); rep(i,1,k+1) scanf("%d",c+i); dfs(1,1,n*m); printf("Case #%d:\n",tt); if(flag) { printf("YES"); rep(i,1,n+1) { rep(j,1,m) printf("%d ",dp[i][j]); printf("%d\n",dp[i][m]); } } else printf("YES"); } return 0; }
HDU 5113 Black And White