标签:get tchar getc == mat 单调队列 str min lin
二维RMQ问题模板。。。(虽然我用单调队列加一维RMQ过的,,,但这个更方便一些,就再拿出来写写。。跑的还快一点。。)
和一维差不多,只不过需要比较四个区域罢了。。
g[i][j][k]=max(g[i][j][k-1],max(g[i+(1<<(k-1))][j][k-1],max(g[i][j+(1<<(k-1))][k-1],g[i+(1<<(k-1))][j+(1<<(k-1))][k-1])));
额,就是这样。。
上代码。
#include<iostream> #include<cstdio> #include<climits> #include<cmath> using namespace std; int a,b,n,t[1010][1010],g[1050][1050][11],h[1050][1050][11],lg; int ans=INT_MAX; inline int read(){ char c=getchar();int x=0,flag=1; while(c<‘0‘ || c>‘9‘){if(c==‘-‘) flag=-1;c=getchar();} while(c>=‘0‘ && c<=‘9‘) x=(x<<1)+(x<<3)+c-‘0‘,c=getchar(); return x*flag; } int main(){ a=read();b=read();n=read(); for(int i=1;i<=a;i++) for(int j=1;j<=b;j++) t[i][j]=read(),g[i][j][0]=h[i][j][0]=t[i][j]; for(int k=1;k<=10;k++){ for(int i=1;i+(1<<k)-1<=a;i++){ for(int j=1;j+(1<<k)-1<=b;j++){ g[i][j][k]=max(g[i][j][k-1],max(g[i+(1<<(k-1))][j][k-1],max(g[i][j+(1<<(k-1))][k-1],g[i+(1<<(k-1))][j+(1<<(k-1))][k-1]))); h[i][j][k]=min(h[i][j][k-1],min(h[i+(1<<(k-1))][j][k-1],min(h[i][j+(1<<(k-1))][k-1],h[i+(1<<(k-1))][j+(1<<(k-1))][k-1]))); } } } lg=log(n)/log(2); for(int i=1;i+n-1<=a;i++){ for(int j=1;j+n-1<=b;j++){ int MAX=max(g[i][j][lg],max(g[i+n-1-(1<<lg)+1][j][lg],max(g[i][j+n-1-(1<<lg)+1][lg],g[i+n-1-(1<<lg)+1][j+n-1-(1<<lg)+1][lg]))); int MIN=min(h[i][j][lg],min(h[i+n-1-(1<<lg)+1][j][lg],min(h[i][j+n-1-(1<<lg)+1][lg],h[i+n-1-(1<<lg)+1][j+n-1-(1<<lg)+1][lg]))); ans=min(ans,MAX-MIN); } } printf("%d\n",ans);return 0; }
标签:get tchar getc == mat 单调队列 str min lin
原文地址:https://www.cnblogs.com/SyhAKIOI/p/11706753.html