Hdu4638:group (line segment tree offline processing)

Problem Descriptionthere was n men, every man had an ID (1..N). The their ID is unique. Whose ID is I and i-1 am friends, whose ID is I and i+1 is friends. These n men stand on line. Now we select a 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 a interval of men and want to know there should being how many groups so the value of interval is max.
Inputfirst line was T indicate the case number.
For each case first line is N, m (1<=n, m<=100000) indicate there be n men and M query.
Then a line has n number indicate the ID of the men from left to right.
Next m Line have a number l,r (1<=l<=r<=n), mean we want to know the answer of [l,r].

Outputfor every query output a number indicate there should be what many group so and the sum of value is max.
Sample Input

15 23 1 2 5 41 52 4

Sample Output

12 Test instructions: Give an array, each query l,r, see the interval can be divided into several consecutive series of ideas: Offline processing all queries
#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 rank1const 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&LT;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", &AMP;S[I].L,&AMP;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;}

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Hdu4638:group (line segment tree offline processing)