[暴力 单调栈] BZOJ 2228 [Zjoi2011]礼物(gift)

这个东西 搞法不一

我是这么搞的

沿着z轴的方向看 一定是一些重叠的正方形叠在一起

我们对于每一个z 都对(x,y)求出fx,y 表示最大的以(x,y)为右下角的正方形的大小

然后在z这一维上 大小就是区间最小值 我们用单调栈弄一弄就好了

我写的二分求fx,y 复杂度O(n3logn)

但是实际上用悬线法可以做到O(n3)

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}
inline void read(char *s){
char c=nc(); int len=0;
for (;!(c>='A' && c<='Z');c=nc());
for (;c>='A' && c<='Z';s[++len]=c,c=nc()); s[++len]=0;
}

const int N=155;

int _g[3]

;

int h
,l
,r
;
int sta
,p;

inline int calc(int n){
h[0]=h[n+1]=-1;
p=0; sta[++p]=0;
for (int i=1;i<=n;i++){
while (p && h[i]<=h[sta[p]]) p--;
l[i]=i-sta[p];
sta[++p]=i;
}
p=0; sta[p]=n+1;
for (int i=n;i;i--){
while (p && h[i]<=h[sta[p]]) p--;
r[i]=sta[p]-i;
sta[++p]=i;
}
int ret=0;
for (int i=1;i<=n;i++)
ret=max(ret,h[i]*(l[i]+r[i]-1));
return ret;
}

int f

;
int g

;
inline int sum(int x1,int y1,int x2,int y2,int k){
return g[k][x2][y2]+g[k][x1-1][y1-1]-g[k][x2][y1-1]-g[k][x1-1][y2];
}

inline int Solve(int p,int q,int r){
for (int k=1;k<=r;k++) for (int i=1;i<=p;i++) for (int j=1;j<=q;j++) g[k][i][j]+=g[k][i][j-1];
for (int k=1;k<=r;k++) for (int i=1;i<=p;i++) for (int j=1;j<=q;j++) g[k][i][j]+=g[k][i-1][j];
for (int k=1;k<=r;k++)
for (int i=1;i<=p;i++)
for (int j=1;j<=q;j++){
int L=0,R=min(i,j)+1,x;
while (L+1<R){
x=(L+R)>>1;
if (!sum(i-x+1,j-x+1,i,j,k))
L=x;
else
R=x;
}
f[k][i][j]=L;
}
int ret=0;
for (int i=1;i<=p;i++)
for (int j=1;j<=q;j++){
for (int k=1;k<=r;k++)
h[k]=f[k][i][j];
ret=max(ret,calc(r));
}
return ret;
}

int main(){
int p,q,r; char str
;
freopen("monument.in","r",stdin);
freopen("monument.out","w",stdout);
read(q); read(p); read(r);
for (int i=1;i<=p;i++)
for (int j=1;j<=q;j++){
read(str);
for (int k=1;k<=r;k++)
_g[0][k][i][j]=_g[1][j][i][k]=_g[2][i][j][k]=str[k]=='P';
}
int Ans=0;
memcpy(g,_g[0],sizeof(g));
Ans=max(Ans,Solve(p,q,r));
memcpy(g,_g[1],sizeof(g));
Ans=max(Ans,Solve(p,r,q));
memcpy(g,_g[2],sizeof(g));
Ans=max(Ans,Solve(q,r,p));
printf("%d\n",Ans*4);
return 0;
}