【HAOI 2007】 理想的正方形

【题目链接】

点击打开链接

【算法】

单调队列

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1010
const int INF = 2e9;

int i,j,a,b,n,ans = INF;
deque<int> q1,q2;
int val[MAXN][MAXN],mx[MAXN][MAXN],mn[MAXN][MAXN];

template <typename T> inline void read(T &x)
{
    int f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
template <typename T> inline void write(T x)
{
    if (x < )
    {
        putchar('-');
        x = -x;
    }
    if (x > ) write(x/);
    putchar(x%+'');
}
template <typename T> inline void writeln(T x)
{
    write(x);
    puts("");
}

int main() {

        read(a); read(b); read(n);
        for (i = ; i <= a; i++)
        {
                for (j = ; j <= b; j++)
                {
                        read(val[i][j]);
                }
        }
        for (i = ; i <= b; i++)
        {
                q1.clear(); q2.clear();
                for (j = ; j < n; j++)
                {
                        while ((!q1.empty()) && (val[q1.back()][i] <= val[j][i])) q1.pop_back();
                        q1.push_back(j);
                        while ((!q2.empty()) && (val[q2.back()][i] >= val[j][i])) q2.pop_back();
                        q2.push_back(j);
                }
                for (j = n; j <= a; j++)
                {
                        while ((!q1.empty()) && (j - q1.front() >= n)) q1.pop_front();
                        while ((!q2.empty()) && (j - q2.front() >= n)) q2.pop_front();
                        while ((!q1.empty()) && (val[q1.back()][i] <= val[j][i])) q1.pop_back();
                        q1.push_back(j);
                        while ((!q2.empty()) && (val[q2.back()][i] >= val[j][i])) q2.pop_back();
                        q2.push_back(j);
                        mx[j][i] = val[q1.front()][i];
                        mn[j][i] = val[q2.front()][i];
                }
        }
        for (i = n; i <= a; i++)
        {
                q1.clear(); q2.clear();
                for (j = ; j < n; j++)
                {
                        while ((!q1.empty()) && (mx[i][q1.back()] <= mx[i][j])) q1.pop_back();
                        q1.push_back(j);
                        while ((!q2.empty()) && (mn[i][q2.back()] >= mn[i][j])) q2.pop_back();
                        q2.push_back(j);
                }
                for (j = n; j <= b; j++)
                {
                        while ((!q1.empty()) && (j - q1.front() >= n)) q1.pop_front();
                        while ((!q2.empty()) && (j - q2.front() >= n)) q2.pop_front();
                        while ((!q1.empty()) && (mx[i][q1.back()] <= mx[i][j])) q1.pop_back();
                        q1.push_back(j);
                        while ((!q2.empty()) && (mn[i][q2.back()] >= mn[i][j])) q2.pop_back();
                        q2.push_back(j);
                        ans = min(ans,mx[i][q1.front()]-mn[i][q2.front()]);
                }
        }
        writeln(ans);

        return ;

}