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AC的不容易啊。。发个帖庆祝下,附代码Source Code
Problem: 1069 User: yzhw
Memory: 320K Time: 79MS
Language: GCC Result: Accepted
Source Code
# include <stdio.h>
# include <stdlib.h>
# include <stdbool.h>
bool map[52][52];
int num,n,list[11];
void printmap()
{
int i,j;
for(i=1;i<=2*n;i++)
{
for(j=1;j<4*n;j++) printf("%d ",map[i][j]);
printf("\n");
}
printf("\n\n");
}
bool canput(int r,int c,int len)
{
int i,j;
if(r+len-1>2*n||c+len-1>4*n) return 0;
if(c%2==0)
{
for(i=r;i<r+len;i++)
for(j=c+2*(i-r);j<c+2*(i-r)+2*len-1-2*(i-r);j++)
if(map[i][j]) return 0;
}
else
{
for(i=r;i<r+len;i++)
for(j=c;j<c+2*(i-r+1)-1;j++)
if(map[i][j]) return 0;
}
return 1;
}
void put(int r,int c,int len)
{
int i,j;
if(c%2==0)
{
for(i=r;i<r+len;i++)
for(j=c+2*(i-r);j<c+2*(i-r)+2*len-1-2*(i-r);j++)
map[i][j]=1;
}
else
{
for(i=r;i<r+len;i++)
for(j=c;j<c+2*(i-r+1)-1;j++)
map[i][j]=1;
}
}
void deput(int r,int c,int len)
{
int i,j;
if(c%2==0)
{
for(i=r;i<r+len;i++)
for(j=c+2*(i-r);j<c+2*(i-r)+2*len-1-2*(i-r);j++)
map[i][j]=0;
}
else
{
for(i=r;i<r+len;i++)
for(j=c;j<c+2*(i-r+1)-1;j++)
map[i][j]=0;
}
}
bool detect(int r,int c)
{
if(r>2*n) return 1;
else if(c>4*n) return detect(r+1,1);
else if(map[r][c])
{
int i;
for(i=c;i<=4*n;i++)
if(!map[r][i]) break;
return detect(r,i);
}
else
{
int i;
for(i=1;i<=num;i++)
{
if(canput(r,c,list[i]))
{
put(r,c,list[i]);
if(detect(r,c+1)) return 1;
deput(r,c,list[i]);
}
else break;
}
return 0;
}
}
int cmp(const void *a,const void *b)
{
return *(int *)a-*(int *)b;
}
int main()
{
int test,i,k;
scanf("%d",&test);
for(i=1;i<=test;i++)
{
int j;
bool flag=0;
scanf("%d",&n);
scanf("%d",&num);
for(j=1;j<=num;j++) scanf("%d",&list[j]);
qsort(list+1,num,sizeof(int),cmp);
for(j=1;j<=num;j++)/*优化*/
{
if(n%list[j]==0)
{
printf("YES\n");
flag=1;
break;
}
if(list[j]>n)
{
num=j-1;
break;
}
}
if(flag) continue;
for(j=1;j<=num;j++)
for(k=1;k<j;k++)
{
if(list[j]%list[k]==0)
{
int l;
for(l=j;l<num;l++) list[l]=list[l+1];
j--;
num--;
break;
}
}
/*---------------------------------------------*/
memset(map,1,sizeof(map));
for(j=1;j<=n;j++)
for(k=1;k<=2*n+1+(j-1)*2;k++)
map[j][k]=0;
for(j=n+1;j<=2*n;j++)
for(k=(j-n)*2;k<=4*n;k++)
map[j][k]=0;
if(detect(1,1)) printf("YES\n");
else printf("NO\n");
}
return 0;
}
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