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  • BZOJ 2724: [Violet 6]蒲公英

    题目:http://www.lydsy.com/JudgeOnline/problem.php?id=2724

    分块+二分。 首先离散化,维护ans[i][j]表示第i块到第j块之间的众数,然后对于询问l,r,在完整块的我们拿出来,然后暴力扫非完整块。

    我们只要把相同的数排在一起,对于a[i],我们可以通过二分找到第一个在r之前的数的位置p和在第一个在l之后的数的位置q,p-q+1即为a[i]在询问l,r的出现次数。然后更新答案就好了。

    (因为函数名起的太丑+手残傻逼了好久TAT。。。

    #include<cstring>
    #include<cctype>
    #include<cstdio>
    #include<algorithm>
    #include<cmath>
    #define rep(i,l,r) for (int i=l;i<=r;i++)
    #define down(i,l,r) for (int i=l;i>=r;i--)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define ll long long
    #define maxn 50005
    #define mm int(1e9+7)
    using namespace std;
    struct data{int v,p;
    }d[maxn];
    int h[maxn],a[maxn],L[maxn],R[maxn],l[maxn],r[maxn],c[maxn],belong[maxn],ans[205][205];
    int n,m,cnt,block,lastans,x,y;
    int read(){
        int x=0,f=1; char ch=getchar();
        while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
        while (isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();}
        return x*f;
    }
    int Find(int x){
        int l=1,r=n;
        while (l<=r){
            int mid=(l+r)/2;
            if (x==h[mid]) return mid;
            if (x<h[mid]) r=mid-1;
            else l=mid+1;
        }
        return l;
    }
    bool cmp(data a,data b){
        return a.v==b.v?a.p<b.p:a.v<b.v;
    }
    void pre(){
        rep(i,1,n) {
            d[i].v=a[i]; d[i].p=i;
        }
        sort(d+1,d+1+n,cmp);
        rep(i,1,n) {
            if (!L[d[i].v]) L[d[i].v]=i;
            R[d[i].v]=i;
        }
        rep(i,1,cnt){
            int tmp,mx=0; 
            clr(c,0);
            rep(j,l[i],n){
                c[a[j]]++;
                if (c[a[j]]>mx||(c[a[j]]==mx&&a[j]<tmp)) mx=c[a[j]],tmp=a[j];
                ans[i][belong[j]]=tmp;
            }
        } 
    }
    int findup(int l,int r,int x){
        int mid,tmp;
        while (l<=r){
            mid=(l+r)/2;
            if (d[mid].p>x) r=mid-1;
            else {tmp=mid; l=mid+1;}
        }
        return tmp;
    }
    int finddown(int l,int r,int x){
        int mid,tmp;
        while (l<=r){
            mid=(l+r)/2;
            if (d[mid].p<x) l=mid+1;
            else {tmp=mid; r=mid-1;} 
        }
        return tmp;
    }
    int que(int x,int y){
        int mx=0,tmp;
        if (belong[x]==belong[y]){
            rep(i,x,y) {
                int num=findup(L[a[i]],R[a[i]],y)-finddown(L[a[i]],R[a[i]],x)+1;
                if (num>mx||(num==mx&&a[i]<tmp)) mx=num,tmp=a[i];
            }
        }
        else {
            if (belong[x]+1<belong[y]){
                tmp=ans[belong[x]+1][belong[y]-1];
                mx=findup(L[tmp],R[tmp],y)-finddown(L[tmp],R[tmp],x)+1;
            }
            rep(i,x,r[belong[x]]){
                int num=findup(L[a[i]],R[a[i]],y)-finddown(L[a[i]],R[a[i]],x)+1;
                if (num>mx||(num==mx&&a[i]<tmp)) mx=num,tmp=a[i];
            }
            rep(i,l[belong[y]],y){
                int num=findup(L[a[i]],R[a[i]],y)-finddown(L[a[i]],R[a[i]],x)+1;
                if (num>mx||(num==mx&&a[i]<tmp)) mx=num,tmp=a[i];
            }
        }
        return tmp;
    }
    int main(){
           n=read(); m=read();
           rep(i,1,n) a[i]=read(),h[i]=a[i];
           sort(h+1,h+1+n);
           rep(i,1,n) a[i]=Find(a[i]);
        block=int(sqrt(n));
        cnt=n/block; if (n%block) cnt++;
        rep(i,1,cnt) l[i]=block*(i-1)+1,r[i]=block*i;
        int now=1; rep(i,1,n) {belong[i]=now;if (i%block==0) now++;}
        pre();
        lastans=0;
        rep(i,1,m){
            x=read(); y=read();
            x=(x+lastans-1)%n+1; y=(y+lastans-1)%n+1;
            if (x>y) swap(x,y);
            printf("%d
    ",lastans=h[que(x,y)]);
        }
        return 0; 
    }
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  • 原文地址:https://www.cnblogs.com/ctlchild/p/4992868.html
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