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;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e8+7;
const double esp = 1e-8;
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=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/
            const int maxn = 1009;
            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=1; i<=n; i++){
                for(int j=1; j<=m; j++){
                    scanf("%d", &mp[i][j]);
                }
            }

            for(int i=1; i<=n; i++){
                    qmx.clear();
                    qmn.clear();
                    for(int j=1; 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() + 1 > 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() + 1 > k) qmn.pop_front();
                        mn[i][j] = mp[i][qmn.front()];
                    }
            }
            int ans = inf;
            for(int j=1; j<=m; j++){
                    qmx.clear();
                    qmn.clear();
                for(int i=1; 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() + 1 > 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() + 1 > 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 0;
}