# 洛谷 P3567 [POI2014]KUR-Couriers

## 题目描述

Byteasar works for the BAJ company, which sells computer games.

The BAJ company cooperates with many courier companies that deliver the games sold by the BAJ company to its customers.

Byteasar is inspecting the cooperation of the BAJ company with the couriers.

He has a log of successive packages with the courier company that made the delivery specified for each package.

He wants to make sure that no courier company had an unfair advantage over the others.

If a given courier company delivered more than half of all packages sent in some period of time, we say that it dominated in that period.

Byteasar wants to find out which courier companies dominated in certain periods of time, if any.

Help Byteasar out!

Write a program that determines a dominating courier company or that there was none.

## 输入输出格式

The first line of the standard input contains two integers,  and  (), separated by a single space, that are the number of packages shipped by the BAJ company and the number of time periods for which the dominating courier is to be determined, respectively.

The courier companies are numbered from  to (at most) .

The second line of input contains  integers,  (), separated by single spaces;  is the number of the courier company that delivered the -th package (in shipment chronology).

The  lines that follow specify the time period queries, one per line.

Each query is specified by two integers,  and  (), separated by a single space.

These mean that the courier company dominating in the period between the shipments of the -th and the -th package, including those, is to be determined.

In tests worth  of total score, the condition  holds, and in tests worth  of total score .

The answers to successive queries should be printed to the standard output, one per line.

(Thus a total of  lines should be printed.) Each line should hold a single integer: the number of the courier company that dominated in the corresponding time period, or  if there was no such company.

## 输入输出样例

7 5
1 1 3 2 3 4 3
1 3
1 4
3 7
1 7
6 6


1
0
3
0
4


## 解析

 1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<cmath>
6 using namespace std;
7 #define ll long long
8 const int maxn=500005;
9 struct node{
10     node *l,*r;
11     int v;
12 }tree[maxn*30],*root[maxn],*cnt;
13 int n,m,tot,tmp;
14 int a[maxn],r[maxn];
15 int le,ri;
16 //r changed a original
17 node *update(node *lst,int l,int r){
18     node *ret=++cnt;
19     *ret=*lst;
20     ret->v++;
21     if (l==r){
22         return ret;
23     }
24     int mid=l+r>>1;
25     if (tmp<=mid) ret->l=update(ret->l,l,mid);
26     else ret->r=update(ret->r,mid+1,r);
27     return ret;
28 }
29 int query(node *u,node *v,int l,int r){
30     if (l==r){
31         return a[l];
32     }
33     int mid=l+r>>1,p=ri-le+1;
34     int p1=v->l->v-u->l->v;
35     int p2=v->r->v-u->r->v;
36     if (2*p1>p) return query(u->l,v->l,l,mid);
37     else if (2*p2>p) return query(u->r,v->r,mid+1,r);
38     else return 0;
39 }
40 int main(){
41     scanf("%d%d",&n,&m);
42     for (int i=1;i<=n;++i){
43         scanf("%d",&r[i]);
44         a[i]=r[i];
45     }
46     sort(a+1,a+1+n);
47     tot=unique(a+1,a+1+n)-a-1;
48     for (int i=1;i<=n;++i){
49         r[i]=lower_bound(a+1,a+1+tot,r[i])-a;
50     }
51     cnt=root[0]=tree->l=tree->r=tree;
52     for (int i=1;i<=n;++i){
53         tmp=r[i];
54         root[i]=update(root[i-1],1,tot);
55     }
56     while (m--){
57         scanf("%d%d",&le,&ri);
58         printf("%d\n",query(root[le-1],root[ri],1,tot));
59     }
60     return 0;
61 }
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posted @ 2018-02-25 21:13  lonlyn  阅读(184)  评论(0编辑  收藏  举报