洛谷P2216: [HAOI2007]理想的正方形 单调队列优化DP

洛谷P2216

）逼着自己写DP

题意：

　　给定一个带有数字的矩阵，找出一个大小为n*n的矩阵，这个矩阵中最大值减最小值最小。

思路:

　　先处理出每一行每个格子到前面n个格子中的最大值和最小值。然后对每一列求出长度为n的前面算出来的最大值的最大值，前面算出来的最小值的最小值。如果直接做是n的三次方，但是用单调队列优化后就是n方的。

#include <algorithm>

#include  <iterator>

#include  <iostream>

#include   <cstring>

#include   <cstdlib>

#include   <iomanip>

#include    <bitset>

#include    <cctype>

#include    <cstdio>

#include    <string>

#include    <vector>

#include     <stack>

#include     <cmath>

#include     <queue>

#include      <list>

#include       <map>

#include       <set>

#include   <cassert>

using namespace std;

#define lson (l , mid , rt << 1)

#define rson (mid + 1 , r , rt << 1 | 1)

#define debug(x) cerr << #x << " = " << x << "\n";

#define pb push_back

#define pq priority_queue

typedef long long ll;

typedef unsigned long long ull;

//typedef __int128 bll;

typedef pair<ll ,ll > pll;

typedef pair<int ,int > pii;

typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q

//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q

#define fi first

#define se second

//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)

#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行

#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)

#define max3(a,b,c) max(max(a,b), c);

#define min3(a,b,c) min(min(a,b), c);

//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //

const int mod = 1e8+;

const double esp = 1e-;

const double PI=acos(-1.0);

const double PHI=0.61803399;    //黄金分割点

const double tPHI=0.38196601;

template<typename T>

inline T read(T&x){

    x=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

/*-----------------------showtime----------------------*/

            const int maxn = ;

            int mp[maxn][maxn];

            int mx[maxn][maxn],mn[maxn][maxn];

            deque<int>qmx,qmn;

int main(){

            int n,m,k;

            scanf("%d%d%d", &n, &m, &k);

            for(int i=; i<=n; i++){

                for(int j=; j<=m; j++){

                    scanf("%d", &mp[i][j]);

                }

            }

            for(int i=; i<=n; i++){

                    qmx.clear();

                    qmn.clear();

                    for(int j=; j<=m; j++){

                        while(!qmx.empty() &&  mp[i][qmx.back()] <= mp[i][j]) qmx.pop_back();

                        qmx.push_back(j);

                        while(!qmx.empty() && j - qmx.front() +  > k) qmx.pop_front();

                        mx[i][j] = mp[i][qmx.front()];

                        while(!qmn.empty() && mp[i][qmn.back()] >= mp[i][j]) qmn.pop_back();

                        qmn.push_back(j);

                        while(!qmn.empty() && j - qmn.front() +  > k) qmn.pop_front();

                        mn[i][j] = mp[i][qmn.front()];

                    }

            }

            int ans = inf;

            for(int j=; j<=m; j++){

                    qmx.clear();

                    qmn.clear();

                for(int i=; i<=n; i++){

                        while(!qmx.empty() &&  mx[qmx.back()][j] <= mx[i][j]) qmx.pop_back();

                        qmx.push_back(i);

                        while(!qmx.empty() && i - qmx.front() +  > k) qmx.pop_front();

                        int tpmx = mx[qmx.front()][j];

                        while(!qmn.empty() && mn[qmn.back()][j] >= mn[i][j]) qmn.pop_back();

                        qmn.push_back(i);

                        while(!qmn.empty() && i - qmn.front() +  > k) qmn.pop_front();

                        int tpmn = mn[qmn.front()][j];

                        if(i>=k&&j>=k) ans = min(ans, tpmx - tpmn);

                }

            }

            printf("%d\n", ans);

            return ;

}