exp limit nds mean tdi 離線 create ber line

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

題意：給出一個數組。每次查詢l,r。看區間內能分成幾個連續的數列
思路：離線處理全部查詢
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define ls 2*i
#define rs 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 100005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define rank rank1
const int mod = 10007;

struct node
{
    int l,r,val;
} a[N*4],s[N];

int num[N],pos[N],vis[N],ans[N];
int n,m;

int cmp(node a,node b)
{
    return a.r<b.r;
}

void build(int l,int r,int i)
{
    a[i].l = l;
    a[i].r = r;
    a[i].val = 0;
    if(l==r) return;
    int mid = (l+r)/2;
    build(l,mid,ls);
    build(mid+1,r,rs);
}

void updata(int i,int pos,int v)
{
    a[i].val+=v;
    if(a[i].l==a[i].r) return;
    int mid = (a[i].l+a[i].r)/2;
    if(pos<=mid) updata(ls,pos,v);
    else updata(rs,pos,v);
}

int query(int l,int r,int i)
{
    if(a[i].l==l&&a[i].r==r)
    {
        return a[i].val;
    }
    int mid = (a[i].l+a[i].r)/2;
    if(r<=mid) return query(l,r,ls);
    if(l>mid) return query(l,r,rs);
    return query(l,mid,ls)+query(mid+1,r,rs);
}

int main()
{
    int t,i,j,k,cnt;
    scanf("%d",&t);
    while(t--)
    {
        MEM(vis,0);
        scanf("%d%d",&n,&m);
        for(i = 1; i<=n; i++)
        {
            scanf("%d",&num[i]);
            pos[num[i]] = i;
        }
        for(i = 1; i<=m; i++)
        {
            scanf("%d%d",&s[i].l,&s[i].r);
            s[i].val = i;
        }
        sort(s+1,s+1+m,cmp);
        build(1,n,1);
        cnt = 1;
        for(i = 1; i<=n&&cnt<=m; i++)
        {
            updata(1,i,1);
            vis[num[i]]=1;
            if(vis[num[i]-1]) updata(1,pos[num[i]-1],-1);
            if(vis[num[i]+1]) updata(1,pos[num[i]+1],-1);
            while(s[cnt].r==i&&cnt<=m)
            {
                ans[s[cnt].val] = query(s[cnt].l,s[cnt].r,1);
                cnt++;
            }
        }
        for(i = 1; i<=m; i++)
            printf("%d\n",ans[i]);
    }

    return 0;
}
 

HDU4638:Group(線段樹離線處理)