【单调队列】BZOJ1047(HAOI2007)[理想的正方形]题解

题目概述

有一个 a×b 的整数组成的矩阵，从中找出一个 n×n 的正方形区域，使得该区域所有数中的最大值和最小值的差最小。

解题报告

水博客……解法很多，可以二维ST表，线段树套线段树（雾）……

O(n) 解法好……先按照行做单调队列预处理出每个点向左推 n 个的极值，然后按照列做单调队列求出每个点向上向左推 n 个的极值就行了。

示例程序

#include<cstdio>#include<cctype>#include<algorithm>using namespace std;const int maxn=1000,MAXINT=((1<<30)-1)*2+1;int n,m,K,pic[maxn+5][maxn+5],QX[maxn+5],QN[maxn+5];int MAX[maxn+5][maxn+5],MIN[maxn+5][maxn+5],ans;#define Eoln(x) ((x)==10||(x)==13||(x)==EOF)inline char readc(){    static char buf[100000],*l=buf,*r=buf;    if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);    if (l==r) return EOF;return *l++;}inline int readi(int &x){    int tot=0,f=1;char ch=readc(),lst='+';    while (!isdigit(ch)) {if (ch==EOF) return EOF;lst=ch;ch=readc();}    if (lst=='-') f=-f;    while (isdigit(ch)) tot=(tot<<3)+(tot<<1)+ch-48,ch=readc();    return x=tot*f,Eoln(ch);}int main(){    freopen("program.in","r",stdin);    freopen("program.out","w",stdout);    readi(n);readi(m);readi(K);ans=MAXINT;    for (int i=1;i<=n;i++)    for (int j=1;j<=m;j++)        readi(pic[i][j]);    for (int i=1;i<=n;i++)    {        int HX=1,TX=0,HN=1,TN=0;        for (int j=1;j<K;j++)        {            while (HX<=TX&&pic[i][QX[TX]]<=pic[i][j]) TX--;            while (HN<=TN&&pic[i][QN[TN]]>=pic[i][j]) TN--;            QX[++TX]=j;QN[++TN]=j;        }        for (int j=K;j<=m;j++)        {            while (HX<=TX&&QX[HX]<=j-K) HX++;            while (HN<=TN&&QN[HN]<=j-K) HN++;            while (HX<=TX&&pic[i][QX[TX]]<=pic[i][j]) TX--;            while (HN<=TN&&pic[i][QN[TN]]>=pic[i][j]) TN--;            QX[++TX]=j;QN[++TN]=j;            MAX[i][j]=pic[i][QX[HX]];MIN[i][j]=pic[i][QN[HN]];        }    }    for (int j=K;j<=m;j++)    {        int HX=1,TX=0,HN=1,TN=0;        for (int i=1;i<K;i++)        {            while (HX<=TX&&MAX[QX[TX]][j]<=MAX[i][j]) TX--;            while (HN<=TN&&MIN[QN[TN]][j]>=MIN[i][j]) TN--;            QX[++TX]=i;QN[++TN]=i;        }        for (int i=K;i<=n;i++)        {            while (HX<=TX&&QX[HX]<=i-K) HX++;            while (HN<=TN&&QN[HN]<=i-K) HN++;            while (HX<=TX&&MAX[QX[TX]][j]<=MAX[i][j]) TX--;            while (HN<=TN&&MIN[QN[TN]][j]>=MIN[i][j]) TN--;            QX[++TX]=i;QN[++TN]=i;            ans=min(ans,MAX[QX[HX]][j]-MIN[QN[HN]][j]);        }    }    return printf("%d\n",ans),0;}