ZJOI 2018 胖

(⊙o⊙)…

 SOL： 我们显然可以二分，我们考虑二分每一个由城堡出来的路径的贡献，必然是一条线段。

  用ST表check就好了。

 

#include<bits/stdc++.h>
#define sight(x) ('0'<=x&&x<='9')
#define N 400017
#define LL long long
#define int LL
using namespace std;
inline void read(int &x) {
    static int b; static char c;
    for (b=1,c=getchar();!sight(c);c=getchar()) if(c=='-') b=-1;
    for (x=0;sight(c);c=getchar()) x=x*10+c-48;
    x=x*b;
}
void write(LL x) {
    if (x<10) {putchar(48+x); return;} write(x/10); putchar(48+x%10);
} 
inline void writeln(LL x) {
    if (x<0) {putchar('-'); x=-x;} write(x); putchar('\n');
}
struct Node{
    int a,b;
    Node() {}
    Node(int x,int y):a(x),b(y){}
    inline bool operator <(const Node& XX)const&{
        return a<XX.a;
    }
}p[N];
#define SIZ 19
#define orz1(x) (1ll*p[x].b+sum[p[x].a])
#define orz2(x) (1ll*p[x].b+sum[n]-sum[p[x].a])
int k,ch1[N][SIZ],ch2[N][SIZ],n,len,b[N],idl,idr;
LL sum[N],key1[N],key2[N];
void build() {
    for (int i=1;i<=k;i++) ch1[i][0]=ch2[i][0]=i,
    key1[i]=1ll*p[i].b+sum[p[i].a],key2[i]=1ll*p[i].b+sum[n]-sum[p[i].a];
    for (int j=1;j<SIZ;j++) 
     for (int i=1;i<=k;i++) {
         if (key1[ch1[i][j-1]]>key1[ch1[i+(1<<j-1)][j-1]]) ch1[i][j]=ch1[i+(1<<j-1)][j-1];
           else ch1[i][j]=ch1[i][j-1]; 
        if (key2[ch2[i][j-1]]<key2[ch2[i+(1<<j-1)][j-1]]) ch2[i][j]=ch2[i][j-1];
           else ch2[i][j]=ch2[i+(1<<j-1)][j-1];

     }
}
inline int que2(int l,int r) {
    len=b[r-l+1];
    return key2[ch2[l][len]]<key2[ch2[r-(1<<len)+1][len]]?ch2[l][len]:ch2[r-(1<<len)+1][len];
}
inline int que1(int l,int r) {
    len=b[r-l+1];
    return key1[ch1[l][len]]>key1[ch1[r-(1<<len)+1][len]]?ch1[r-(1<<len)+1][len]:ch1[l][len];
}
int p1,p2,anw;
inline bool check1(int x,int id) {
    if (x<=0) return 0;
    idl=lower_bound(p+1,p+k+1,Node(x,0))-p;
    idr=lower_bound(p+1,p+k+1,Node(x*2-p[id].a,0))-p;
    if (idr<idl) p1=que2(idr,idl-1);
    else p1=0;  
    p2=que1(idl,id);
    if (!p1) anw=p2; else if (!p2) anw=p1; else 
    if (p[p1].b+sum[x]-sum[p[p1].a]!=p[p2].b+sum[p[p2].a]-sum[x])
       anw=(p[p1].b+sum[x]-sum[p[p1].a]>p[p2].b+sum[p[p2].a]-sum[x])?p2:p1;
    else anw=p[p2].a-x>x-p[p1].a?p1:p2;
    return anw==id;
}
inline bool check2(int x,int id) {
    if (x>n) return 0;
    idl=lower_bound(p+1,p+k+1,Node(x,0))-p;
    idr=upper_bound(p+1,p+k+1,Node(x*2-p[id].a,0))-p;
    p1=que2(id,idl-1);  
    if (idl<idr) p2=que1(idl,idr-1);
    else p2=0;
    if (!p1) anw=p2; else if (!p2) anw=p1; else 
    if (p[p1].b+sum[x]-sum[p[p1].a]!=p[p2].b+sum[p[p2].a]-sum[x])
     anw=(p[p1].b+sum[x]-sum[p[p1].a]>p[p2].b+sum[p[p2].a]-sum[x])?p2:p1;
    else anw=p[p2].a-x>x-p[p1].a?p1:p2;
    return anw==id;
}
int m,x,l,r,siz;
LL ans;
signed main () {
    for (int i=2;i<N;i++) b[i]=b[i>>1]+1;
    read(n); read(m);
    for (int i=2;i<=n;i++) read(x),sum[i]=sum[i-1]+x;
    while (m--) {
        read(k); ans=0;
        for (int i=1;i<=k;i++) 
         read(p[i].a),read(p[i].b);
        sort(p+1,p+k+1);
        build();
        for (int i=1;i<=k;i++) {
//            if (i==50) {
//                l=0;
//            }
            siz=1<<18; l=r=p[i].a;                                                                                                                                                                                                                                                                                    // 1+1=fhbhandsome
            while (siz) {
                if (check1(l-siz,i)) l-=siz;
                if (check2(r+siz,i)) r+=siz;
                siz>>=1; 
            }
            ans+=r-l+1;
//           if (m==98)
//           cerr<<i<<' '<<l<<' '<<r<<endl;
        }
       writeln(ans);
    } 
    return 0;
}