HDU 5113 Black And White(DFS+剪枝)
題面:
Black And White
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1336 Accepted Submission(s): 350
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 c
i 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 c
i (c
i > 0), denoting the number of cells where the i-th color should be used.
It’s guaranteed that c
1 + c
2 + · · · + c
K = 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
Source 2014ACM/ICPC亞洲區北京站
解題:
因為n,m比較小,會想到搜索,但直接超時了。需要加一個剪枝,因為當剩下的容量為n時,任意一種顏色最多只能為(n+1)/2。當任意一種顏色數量超過這個值時,就返回。這個優化已經很厲害了。我又加了一個沒什麼用的優化,main函數中出現某兩種顏色相同時,不用重復計算。
代碼:
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
using namespace std;
int map [7][7],color[30];
int n,m,t,k,sz,amount,tmp;
bool flag=false;
void init()
{
for(int i=0;i<7;i++)
map[0][i]=-1;
for(int j=1;j<7;j++)
map[j][0]=-1;
}
void dfs(int x,int y,int typ,int lef)
{
tmp=(lef+1)/2;
for(int i=1;i<=k;i++)
{
if(color[i]>tmp)
return;
}
if(flag)return;
if(color[typ]==0)return;
if(map[x-1][y]!=typ&&map[x][y-1]!=typ)
{
map[x][y]=typ;
color[typ]--;
if(x==n&&y==m)
{
flag=true;
return;
}
else if(y==m)
{
tmp=0;
for(int i=1;i<=k;i++)
{
if(color[i])tmp++;
}
if(tmp==1)
{
if(m!=1)
{
color[typ]++;
return;
}
}
for(int i=1;i<=k;i++)
{
if(color[i])
dfs(x+1,1,i,lef-1);
}
}
else
{
for(int i=1;i<=k;i++)
{
if(color[i])
{
dfs(x,y+1,i,lef-1);
}
}
}
color[typ]++;
}
return;
}
int main()
{
init();
scanf(%d,&t);
for(int i=1;i<=t;i++)
{
set cnt;
printf(Case #%d:
,i);
scanf(%d%d%d,&n,&m,&k);
amount=(n*m+1)/2;
flag=true;
if(k==1)
{
if(n*m!=1)
flag=false;
}
for(int j=1;j<=k;j++)
{
scanf(%d,&color[j]);
if(color[j]>amount)
flag=false;
}
if(flag)
{
flag=false;
for(int j=1;j<=k;j++)
{
sz=cnt.size();
if(color[j])
{
cnt.insert(color[j]);
if(cnt.size()>sz)
dfs(1,1,j,n*m);
}
}
}
if(flag)
{
printf(YES
);
for(int j=1;j<=n;j++)
{
printf(%d,map[j][1]);
for(int k=2;k<=m;k++)
printf( %d,map[j][k]);
printf(
);
}
}
else
{
printf(NO
);
}
}
return 0;
}