HNOI2017影魔

影魔

• 这么简单的方法尽然想不到，我是真的菜
• 对每个点，用单调栈的方式处理出他左右第一个比他大的数的位置，你可以把(0)和(n+1)设成(inf)。
• 显然对于每对(lef[i])和(rig[i])都会做出(p1)的贡献
• 每个(lef[i])会对(i+1)到(rig[i]-1)做出(p2)贡献
• 同理，每个(rig[i])都会给(lef[i]+1)到(i-1)做出(p2)贡献
• 用结构体存下来，按顺序用线段树将贡献加入即可
• 统计贡献，对于每个询问(l-r)在扫到(l-1)时将这段区间减去，在扫到(r)时再将这段区间贡献加上即可

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
typedef int sign;
typedef long long ll;
#define For(i,a,b) for(register sign i=(sign)a;i<=(sign)b;++i)
#define Fordown(i,a,b) for(register sign i=(sign)a;i>=(sign)b;--i)
const int N=2e5+5;
bool cmax(sign &a,sign b){return (a<b)?a=b,1:0;}
bool cmin(sign &a,sign b){return (a>b)?a=b,1:0;}
template<typename T>inline T read()
{
    T f=1,ans=0;
    char ch=getchar();
    while(!isdigit(ch)&&ch!='-')ch=getchar();
    if(ch=='-')f=-1,ch=getchar();
    while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch-'0'),ch=getchar();
    return ans*f;
}
template<typename T>inline void write(T x,char y)
{
    if(x==0)
    {
        putchar('0');putchar(y);
        return;
    }
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    static char wr[20];
    int top=0;
    for(;x;x/=10)wr[++top]=x%10+'0';
    while(top)putchar(wr[top--]);
    putchar(y);
}
void file()
{
    #ifndef ONLINE_JUDGE
        freopen("3722.in","r",stdin);
        freopen("3722.out","w",stdout);
    #endif
}
int n,m,p1,p2;
int a[N];
struct Q
{
    int l,r,x,tag,id;
    bool operator < (const Q &s)const {return x<s.x;} 
}opt[N<<2];
int cnt;
ll ans[N];
void input()
{
    int l,r;
    n=read<int>();m=read<int>();p1=read<int>();p2=read<int>();
    For(i,1,n)a[i]=read<int>();
    For(i,1,m)
    {
        l=read<int>();r=read<int>();
        ans[i]+=1ll*(r-l)*1ll*p1;
        opt[++cnt]=(Q){l,r,l-1,-1,i};
        opt[++cnt]=(Q){l,r,r,1,i};
    }
}
const int inf=0x3f3f3f3f;
namespace Tree
{
    #define mid ((l+r)>>1)
    #define lson h<<1,l,mid
    #define rson h<<1|1,mid+1,r
    ll sum[N<<2],lazy[N<<2];
    void push_up(int h)
    {
        sum[h]=sum[h<<1]+sum[h<<1|1];
    }
    void push_down(int h,int l,int r)
    {
        if(!lazy[h])return;
        int ls=h<<1,rs=ls|1;
        lazy[ls]+=lazy[h];lazy[rs]+=lazy[h];
        sum[ls]+=lazy[h]*1ll*(mid-l+1);
        sum[rs]+=lazy[h]*1ll*(r-mid);
        lazy[h]=0;
    }
    void update(int h,int l,int r,int s,int t,ll v)
    {
        if(s<=l&&r<=t)
        {
            lazy[h]+=v;
            sum[h]+=1ll*v*1ll*(r-l+1);
        }
        else
        {
            push_down(h,l,r);
            if(s<=mid)update(lson,s,t,v);
            if(mid<t)update(rson,s,t,v);
            push_up(h);
        }
    }
    ll query(int h,int l,int r,int s,int t)
    {
        if(s<=l&&r<=t)return sum[h];
        push_down(h,l,r);
        ll res=0;
        if(s<=mid)res=query(lson,s,t);
        if(mid<t)res+=query(rson,s,t);
        push_up(h);
        return res;
    }
}
#define rg register
int lef[N],rig[N];
struct node
{
    int l,r,x;
    ll v;
    bool operator < (const node &s)const {return x<s.x;}
}e[N<<2];
int sz;
void init()
{
    int j;
    sort(opt+1,opt+cnt+1);
    a[0]=a[n+1]=inf;
    For(i,1,n)
    {
        for(j=i-1;j>=0;j=lef[j])if(a[j]>a[i])break;
        lef[i]=j;
    }
    Fordown(i,n,1)
    {
        for(j=i+1;j<=n+1;j=rig[j])if(a[j]>a[i])break;
        rig[i]=j;
    }
    //For(i,1,n)cout<<lef[i]<<' '<<rig[i]<<endl;
    For(i,1,n)
    {
        if(lef[i]+1<i&&rig[i]<=n)e[++sz]=(node){lef[i]+1,i-1,rig[i],p2};
        if(1<=lef[i]&&rig[i]>i+1)e[++sz]=(node){i+1,rig[i]-1,lef[i],p2};
        if(1<=lef[i]&&rig[i]<=n)e[++sz]=(node){lef[i],lef[i],rig[i],p1};
    }
    sort(e+1,e+sz+1);
}
void work()
{
    int pos1=1,pos2=1;
    while(pos2<=cnt&&!opt[pos2].x)pos2++;
    //cerr<<pos2<<' '<<cnt<<endl;
    For(i,1,n)
    {
        for(;pos1<=sz&&e[pos1].x==i;++pos1)
            Tree::update(1,0,n,e[pos1].l,e[pos1].r,e[pos1].v);
        for(;pos2<=cnt&&opt[pos2].x==i;++pos2)
            ans[opt[pos2].id]+=Tree::query(1,0,n,opt[pos2].l,opt[pos2].r)*opt[pos2].tag;
    }
    For(i,1,m)write(ans[i],'
');
}
int main()
{
    file();
    input();
    init();
    work();
    return 0;
}