CF-1195E

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题意

给一个 $nm(1 \leq n, m \leq 3000)$ 的矩阵，求每一个 $ab$ 的子矩阵中最小值之和

题解

先求出每一行长度为 $b$ 的序列的最小值，再求每一列长度为 $a$ 的序列的最小值

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define memcp(a,b) memcpy(a,b,sizeof(b))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pu_b push_back
#define pu_f push_front
#define po_b pop_back
#define po_f pop_front
#define fi first
#define se second
#define whiel while
#define retrun return
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%Lf",&a);}
void out(int a){printf("%d ",a);}
void outln(int a){printf("%d\n",a);}
void out(ll a){printf("%lld ",a);}
void outln(ll a){printf("%lld\n",a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 3.1e3;
const ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b); }
 
int minn[N][N],h[N][N],q[N];// q[i]表示从 i 开始长度为 len 的序列中的最小值的下标
ll ans;
void fun(int x,int n,int a){
    int l=1,r=0;
    for1(i,n){
        whiel(i-q[l]+1>a&&l<=r)l++;
        while(h[x][q[r]]>h[x][i]&&l<=r)r--;
        q[++r]=i;
        // cout<<i<<' '<<l<<' '<<q[l]<<endl;
        minn[x][i]=h[x][q[l]];
    }
}
void fun2(int x,int n,int a){
    int l=1,r=0;
    for(int i=1;i<=n;i++){
        whiel(i-q[l]+1>a&&l<=r)l++;
        while(minn[q[r]][x]>minn[i][x]&&l<=r)r--;
        q[++r]=i;
        if(i>=a)
        ans+=minn[q[l]][x];
        // minn[x][i]=h[x][q[l]];
    }
}
int main() {
#ifdef PerpEternal
    // freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n,m,a,b,g,x,y,z;
    in(n,m,a,b);
    in(g,x,y,z);
    for0(i,n*m){
        h[i/m+1][i%m+1]=g;
        g=(1ll*g*x+y)%z;
    }
    // for1(i,n){
    //     for1(j,m){
    //         out(h[i][j]);
    //     }
    //     puts("");
    // }
    for1(i,n){
        fun(i,m,b);
    }
    // for1(i,m)out(minn[1][i]);
    // puts("");
    for(int i=b;i<=m;i++){
        fun2(i,n,a);
    }
    outln(ans);
    return 0;
}