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zc
2019-12-21 19:47:00
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先把所有区间按右端点排序。

因为m \le 10,所以直接枚举k值就可以了

假设当前位置为i,现在要找位置j满足|b_i-b_j|等于当前k值,可以是b_i-b_j,即b_i > k,也可以是b_j - b_i = k

b_i+k = b_j, b_i+k\le n我们事先用p_x记录值为xb数组中的哪个位置(b_i互不相同,也就是b数组是1~n的组合)

用线段树维护位置i可以达到的最大的值就可以了,因为已经按右端点排序好了,不会超过当前区间的右端点

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;i++)
    #define Fdr(i,x,y) for(int i=x;i>=y;i--)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 100011
int n,m,Q,a[N],b[N],k[N],ans[N],p[N],s[N<<2];
#define ls rt<<1
#define rs rt<<1|1
#define pu s[rt]=MAX(s[ls],s[rs])
void upd(int x,int c,int l=1,int r=n,int rt=1){
    if(l==r){
        s[rt]=MAX(s[rt],c);
        return;
    }
    int m=(l+r)>>1;
    if(x<=m)upd(x,c,l,m,ls);
    else upd(x,c,m+1,r,rs);
    pu;
}
int ask(int L,int R,int l=1,int r=n,int rt=1){
    if(L<=l&&r<=R)return s[rt];
    int m=(l+r)>>1,ans=-(1<<30);
    if(L<=m)ans=ask(L,R,l,m,ls);
    if(R>m)ans=MAX(ans,ask(L,R,m+1,r,rs));
    return ans;
}
struct que{
    int l,r,id;
}q[N];
bool cmp(que x,que y){
    return x.r<y.r;
}
int main(){
    in>>n>>Q>>m;
    Fur(i,1,n)in>>a[i];
    Fur(i,1,n)in>>b[i],p[b[i]]=i;
    Fur(i,1,m)in>>k[i];
    Fur(i,1,Q)in>>q[i].l>>q[i].r,q[i].id=i;
    sort(q+1,q+Q+1,cmp);
    int x=0;
    Fur(i,1,Q){
        while(x<q[i].r){
            x++;
            Fur(j,1,m){
                if(b[x]>k[j]&&p[b[x]-k[j]]<=x)
                    upd(p[b[x]-k[j]],a[x]+a[p[b[x]-k[j]]]);
                if(b[x]+k[j]<=n&&p[k[j]+b[x]]<=x)
                    upd(p[k[j]+b[x]],a[x]+a[p[k[j]+b[x]]]);
            }
        }
        ans[q[i].id]=ask(q[i].l,q[i].r);
    }
    Fur(i,1,Q)out<<ans[i]<<ln;
}
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