hdu 4638 Group如何用线段树离线维护左边界，实现4级查询？

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题目要求：将以下内容进行简化改写，不超过100字，不使用数字。

简化内容：Group Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submissions: 369 Accepted Submissions: 207 Problem Description: There are n men, each with a unique ID (1..n). Find whose ID is i and i is what?

改写结果：时间限制：Java/Others 4/2秒，内存限制：32/32K。共提交369次，通过207次。题目：有n个男人，每人有唯一ID（1..n），求ID为i的人ID是多少？

Group

Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 369Accepted Submission(s): 207

Problem Description

There are n men ,every man has an ID(1..n).their ID is unique. Whose ID is i and i-1 are friends, Whose ID is i and i+1 are friends. These n men stand in line. Now we select an interval of men to make some group. K men in a group can create K*K value. The value of an interval is sum of these value of groups. The people of same group's id must be continuous. Now we chose an interval of men and want to know there should be how many groups so the value of interval is max.

Input

First line is T indicate the case number. For each case first line is n, m(1<=n ,m<=100000) indicate there are n men and m query. Then a line have n number indicate the ID of men from left to right. Next m line each line has two number L,R(1<=L<=R<=n),mean we want to know the answer of [L,R].

Output

For every query output a number indicate there should be how many group so that the sum of value is max.

Sample Input

15 23 1 2 5 41 52 4

Sample Output

12

Source

​​2013 Multi-University Training Contest 4 ​​

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zhuyuanchen520

思路：离线处理，维护左边界，树状数组记录，从左界到开始的不连续段个数。

如何找不连续段呢？ 判断比i大1小1的再不在区间就可以了。接下来就水了。

失误：多校时怎么没想到这么找区间呢。其实感觉不难。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))using namespace std;const int mm=1e5+9;int c[mm],id[mm],zf[mm],ans[mm];class Edge{public: int l,r,id; bool operator <(const Edge&x)const { return l<x.l; }} f[mm];int lowbit(int x){ return x&(-x);}void add(int pos,int val){ for(int i=pos; i<mm; i+=lowbit(i)) c[i]+=val;}int get_sum(int pos){ int ret=0; for(int i=pos; i>0; i-=lowbit(i)) ret+=c[i]; return ret;}int cas,n,m;int main(){ while(~scanf("%d",&cas)) { while(cas--) { scanf("%d%d",&n,&m); FOR(i,1,n) { scanf("%d",&zf[i]); id[zf[i]]=i; } id[0]=id[n+1]=n+3; FOR(i,1,m) { scanf("%d%d",&f[i].l,&f[i].r);f[i].id=i; } clr(c,0); FOR(i,1,n) { if( id[ zf[i]+1 ]>i && id[ zf[i]-1 ]>i )///两个都在后面是孤立点 { add(i,1); } else if( id[ zf[i]+1 ]<i && id[ zf[i]-1 ]<i )///两个都在前面，加入就合并 { add(i,-1); } } sort(f+1,f+m+1); // FOR(i,1,n)//cout<<c[i]<<endl; //cout<<get_sum(6)<<endl; int i=1,j=1; while(j<=m) { while(i<=n&&i<f[j].l)//去除小于左界影响 { if( id[zf[i]+1 ]>i && id[zf[i]-1 ]>i ) { add(i,-1); add(id[zf[i]+1 ],1); add(id[zf[i]-1 ],1); } else if( id[zf[i]+1 ]<i && id[zf[i]-1 ]<i ) { add(i,-1); } else if( id[zf[i]+1 ]>i ) { add(i,-1); add(id[zf[i]+1],1); } else if( id[zf[i]-1 ]>i ) { add(i,-1); add(id[zf[i]-1 ],1); } ++i; } while(j<=m&&f[j].l<=i) { ans[f[j].id]=get_sum(f[j].r); ++j; } } FOR(i,1,m) printf("%d\n",ans[i]); } } return 0;}