題目鏈接

BZOJ4826

題解

蒟蒻智力水平捉急orz

我們會發現相鄰的\(i\)和\(j\)貢獻一定是\(p1\)，可以很快算出來【然而我一開始忘了考慮調了半天】

我們現在只考慮不相鄰的
我們只需要找出所有產生貢獻的\(i,j\)即可
我們發現每一個產生貢獻的\(i,j\)都能對應到一個三元組\((i,k,j)\)，分別對應區間的最大值，次大值，第三大值
我們枚舉中間位置\(i\)，找到\(i\)左邊第一個比\(i\)大的位置\(L[i]\)，右邊第一個比\(i\)大的位置\(R[i]\)
那麽\(L[i]\)和\(R[i]\)的貢獻就是\(p1\)
區間\((L[i],i)\)和\(R[i]\)

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<vector>
#include<cmath>
#define REP(i,n) for (int i = 1; i <= (n); i++)
 

#define LL long long int
#define ls (u << 1)
#define rs (u << 1 | 1)
using namespace std;
const int maxn = 200005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48
 
 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
struct node{int l,r;};
struct Que{int l,r,pos,t,id;}q[maxn << 1];
vector<node> Ln[maxn],Rn[maxn];
LL ans[maxn];
LL A[maxn],n,m,p1,p2,qi;
LL st[maxn],top,Li[maxn],Ri[maxn];
LL sum[maxn << 2],tag[maxn << 2];
inline bool operator <(const Que& a,const Que& b){
    return a.pos < b.pos;
}
void upd(int u){sum[u] = sum[ls] + sum[rs];}
void pd(int u,int l,int r){
    if (tag[u]){
        int mid = l + r >> 1;
        sum[ls] += tag[u] * (mid - l + 1);
        sum[rs] += tag[u] * (r - mid);
        tag[ls] += tag[u]; tag[rs] += tag[u];
        tag[u] = 0;
    }
}
void add(int u,int l,int r,int L,int R,LL v){
    if (l >= L && r <= R){sum[u] += v * (r - l + 1); tag[u] += v; return;}
    pd(u,l,r);
    int mid = l + r >> 1;
    if (mid >= L) add(ls,l,mid,L,R,v);
    if (mid < R) add(rs,mid + 1,r,L,R,v);
    upd(u);
}
LL query(int u,int l,int r,int L,int R){
    if (l >= L && r <= R) return sum[u];
    pd(u,l,r);
    int mid = l + r >> 1;
    if (mid >= R) return query(ls,l,mid,L,R);
    if (mid < L) return query(rs,mid + 1,r,L,R);
    return query(ls,l,mid,L,R) + query(rs,mid + 1,r,L,R);
}
void init(){
    for (int i = 1; i <= n; i++){
        while (top && A[i] > A[st[top]]) top--;
        Li[i] = st[top];
        st[++top] = i;
    }
    st[0] = n + 1; top = 0;
    for (int i = n; i; i--){
        while (top && A[i] > A[st[top]]) top--;
        Ri[i] = st[top];
        st[++top] = i;
    }
    for (int i = 1; i <= n; i++){
        //printf("pos: %d   L:[%d,%d]  R:[%d,%d]\n",i,Li[i] + 1,i - 1,i + 1,Ri[i] - 1);
        if (Li[i])
            Ln[Li[i]].push_back((node){i + 1,Ri[i] - 1});
        if (Ri[i] <= n)
            Rn[Ri[i]].push_back((node){Li[i] + 1,i - 1});
    }
}
void solve(){
    sort(q + 1,q + 1 + qi); int l,r,pos = 1;
    for (int i = 0; i <= n; i++){
        for (unsigned int j = 0; j < Ln[i].size(); j++){
            l = Ln[i][j].l; r = Ln[i][j].r;
            if (l <= r) add(1,1,n,l,r,p2);
        }
        for (unsigned int j = 0; j < Rn[i].size(); j++){
            l = Rn[i][j].l; r = Rn[i][j].r;
            if (l <= r) add(1,1,n,l,r,p2);
            if (l - 1 > 0) add(1,1,n,l - 1,l - 1,p1);
        }
        while (pos <= qi && q[pos].pos == i){
            ans[q[pos].id] += query(1,1,n,q[pos].l,q[pos].r) * q[pos].t;
            pos++;
        }
    }
    for (int i = 1; i <= m; i++)
        printf("%lld\n",ans[i]);
}
int main(){
    n = read(); m = read(); p1 = read(); p2 = read(); int l,r;
    for (int i = 1; i <= n; i++) A[i] = read();
    for (int i = 1; i <= m; i++){
        l = read(); r = read();
        q[++qi] = (Que){l,r,l - 1,-1,i};
        q[++qi] = (Que){l,r,r,1,i};
        ans[i] += 1ll * (r - l) * p1;
    }
    init();
    solve();
    return 0;
}

BZOJ4826 [Hnoi2017]影魔 【線段樹 + 單調棧】