HDU - 5306 Gorgeous Sequence

传送门：HDU - 5306

思路分析

记录一下区间最大值，最大值数量，次大值
如果当前区间的最大值小于val，就直接返回
如果val大于次大值的话，只有最大值会受到影响
剩下的直接暴力递归下去
查询的时候也要pushdown，因为之前存在区间被修改

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 1e6+10;

struct node {
    int l,r,mid,ma,cnt,sema;
    ll sum;
} tree[N<<2];

int a[N];

#define ls x<<1
#define rs x<<1|1
void pp(int x) {
    tree[x].sum=tree[ls].sum+tree[rs].sum;
    tree[x].ma=max(tree[ls].ma,tree[rs].ma);
    tree[x].sema=max(tree[ls].sema,tree[rs].sema);
    tree[x].cnt=0;
    if(tree[ls].ma!=tree[rs].ma) tree[x].sema=max(tree[x].sema,min(tree[ls].ma,tree[rs].ma));
    if(tree[x].ma==tree[ls].ma) tree[x].cnt+=tree[ls].cnt;
    if(tree[x].ma==tree[rs].ma) tree[x].cnt+=tree[rs].cnt;
}


void build(int x,int l,int r) {
    int mid=l+r>>1;
    tree[x]= {l,r,mid};
    if(l==r) {
        tree[x].cnt=1;
        tree[x].ma=a[l];
        tree[x].sum=a[l];
        tree[x].sema=-INF;
        return ;
    }
    build(ls,l,mid);
    build(rs,mid+1,r);
    pp(x);
}

void solve(int x,int val) {
    if(val<tree[x].ma) {
        tree[x].sum-=1LL*(tree[x].ma-val)*tree[x].cnt;
        tree[x].ma=val;
    }
}

void pd(int x) {
    solve(ls,tree[x].ma);
    solve(rs,tree[x].ma);
}

void modify(int x,int l,int r,int val) {
    if(val>=tree[x].ma) return ;
    if(l<=tree[x].l && tree[x].r<=r && val > tree[x].sema) {
        solve(x,val);
        return ;
    }
    pd(x);
    int mid=tree[x].mid;
    if(l<=mid) modify(ls,l,r,val);
    if(r>mid) modify(rs,l,r,val);
    pp(x);
}

int getmax(int x,int l,int r) {
    if(l<=tree[x].l && tree[x].r<=r) {
        return tree[x].ma;
    }
    pd(x);
    int mid=tree[x].mid;
    int ans=-INF;
    if(l<=mid) ans=max(ans,getmax(ls,l,r));
    if(r>mid) ans=max(ans,getmax(rs,l,r));
    return ans;
}

ll getsum(int x,int l,int r) {
    if(l<=tree[x].l && tree[x].r<=r) {
        return tree[x].sum;
    }
    pd(x);
    ll ans=0;
    int mid=tree[x].mid;
    if(l<=mid) ans+=getsum(ls,l,r);
    if(r>mid) ans+=getsum(rs,l,r);
    return ans;
}
#undef ls
#undef rs

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int T;
    read(T);
    while(T--) {
        int n,m;
        read(n,m);
        for(int i=1; i<=n; i++) read(a[i]);
        build(1,1,n);
        while(m--) {
            int op,l,r,x;
            read(op,l,r);
            if(op==0) {
                read(x);
                modify(1,l,r,x);
            }
            if(op==1) printf("%d\n",getmax(1,l,r));
            if(op==2) printf("%lld\n",getsum(1,l,r));
        }

    }




    return 0;
}