hdu 4368 樹狀陣列 離線維護

life4711發表於2015-08-29
陣列

http://acm.hdu.edu.cn/showproblem.php?pid=4638

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
1
5 2
3 1 2 5 4
1 5
2 4





 



Sample Output




1
2






/**
hdu 4368 樹狀陣列
題目大意：給定1~n的一個排列，每次查詢一個區間(l,r)，將所求區間內的數進行分組，每一組內的數的值必須是連著的，該組的價值為元素個數的平方和
          問在區間價值最大的情況下，要分成幾組
解題思路：用樹狀陣列離線維護，把詢問按照右區間遞增排序，然後從1~n維護，維護到i時注意找尋前i個是不是有a[i]-1,和a[i]+1，有的話直接去掉，因為
          這兩個數必定要和a[i]分到一組，留一個就好了
*/
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
using namespace std;
const int maxn=100100;
int n,m,a[maxn],p[maxn],ans[maxn];
struct note
{
    int l,r,id;
    bool operator < (const note &other)const
    {
        return r<other.r;
    }
}que[maxn];

int c[maxn];

int lowbit(int x)
{
    return x&(-x);
}
int sum(int x)
{
    int ret=0;
    while(x>0)
    {
        ret+=c[x];x-=lowbit(x);
    }
    return ret;
}

void add(int x,int d)
{
    while(x<=n)
    {
        c[x]+=d;
        x+=lowbit(x);
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            p[a[i]]=i;
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&que[i].l,&que[i].r);
            que[i].id=i;
        }
        sort(que,que+m);
        memset(c,0,sizeof(c));
        int j=0;
        for(int i=1;i<=n;i++)
        {
            add(i,1);
            if(a[i]>1&&p[a[i]-1]<i)add(p[a[i]-1],-1);
            if(a[i]<n&&p[a[i]+1]<i)add(p[a[i]+1],-1);
            while(j<m&&que[j].r==i)
            {
                ans[que[j].id]=sum(que[j].r)-sum(que[j].l-1);
                j++;
            }
        }
        for(int i=0;i<m;i++)
        {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}









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