Black And White
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 485 Accepted Submission(s): 131
Special Judge
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.
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.
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
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
技巧性解法：
先对棋盘黑白标记，然后把所有的颜色种类分成小于等于（n*m+1)/2的三类。 a1 > a2 > a3 。无法划分则无解。
然后a1类从上往下填黑色标记的格子，a2类从下往上填白色标记的格子，剩余的格子用a3类填(经过不严格证明 a3类的不会相邻)。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
const int maxn=51;
int i,j,k;
int n,m,nm,nm1,nm2,bj;
int a[maxn],wz1[maxn],ans[maxn];
int b[maxn],wz2[maxn],col[maxn];
int line[4][maxn],gs[4];
int cmp(int x,int y){
return gs[x]>gs[y];
}
int main(){
int T,q;
scanf("%d",&T);
for(int ca=1;ca<=T;++ca){
printf("Case #%d:\n",ca);
scanf("%d%d%d",&n,&m,&q);
for(i=1;i<=q;++i){
scanf("%d",&a[i]);
}
nm=n*m;
nm1=(nm+1)>>1;
nm2=nm>>1;
int l1=0,l2=0;
col[0]=0;
for(int i=2;i<=m;++i) col[i]=1-col[i-1];
for(int i=m+1;i<=nm;++i) col[i]=1-col[i-m];
for(int i=1;i<=nm;++i){
if(col[i])wz2[++l2]=i;
else wz1[++l1]=i;
}
gs[1]=gs[2]=gs[3]=0;
for(i=1;i<4;++i)b[i]=i;
for(i=1;i<=q && gs[1]+a[i]<=nm1;++i){
while(a[i]--)line[1][++gs[1]]=i;
}
for(;i<=q && gs[2]+a[i]<=nm1;++i){
while(a[i]--)line[2][++gs[2]]=i;
}
for(;i<=q && gs[3]+a[i]<=nm1;++i){
while(a[i]--)line[3][++gs[3]]=i;
}
if(i<=q){
puts("NO");
continue;
}
puts("YES");
sort(b+1,b+4,cmp);
for(i=1;i<=nm1-gs[b[1]];++i){
ans[wz1[i]]=line[b[3]][i];
}
k=i;
for(j=1;j<=gs[b[1]];++j,++i){
ans[wz1[i]]=line[b[1]][j];
}
for(i=nm2;i>gs[b[2]];--i,++k){
ans[wz2[i]]=line[b[3]][k];
}
for(;i>0;--i){
ans[wz2[i]]=line[b[2]][i];
}
for(int i=0;i<n;++i){
for(int j=0;j<m;++j){
if(j)printf(" ");
printf("%d",ans[i*m+1+j]);
}
puts("");
}
}
// system("pause");
return 0;
}