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],‘\n‘);
  }
  int main()
  {
  file();
  input();
  init();
  work();
  return 0;
  }

原文地址：https://www.cnblogs.com/dengyixuan/p/8361947.html