HDU 5316 segment tree interval maximum value problem

T-Group data

n Numbers, M operations

Op=0: Find the maximum ' value ' in the L-r interval

Op=1: Change the number of a position to B

Definition of maximum ' value ': the maximal sub-sequence within the interval needs to be guaranteed that the subscript of the sub-sequence is an odd-even alternating

Each interval is recorded with a segment tree, respectively.

EE: The maximum and the end of even subscript starting with even subscript

EO: The most large and the end of an odd subscript starting with even subscript

OE: The most large and even subscript ends with an odd subscript

OO: The most large and large number of subscript starts with odd subscript end

The maximum number of 4 cases to be processed per query

Modify a single point of update can

#include "stdio.h" #include "string.h" const __int64 inf=0x3f3f3f3f3f3f3f3f;__int64 ans,a[100100];struct node{int l,r; __int64 Ee,eo,oe,oo;} Data[400010];__int64 Max (__int64 A,__int64 b) {if (a<b) return b;else return A;    void pushup (int k) {Data[k].ee=max (-inf,max (data[k*2].ee+data[k*2+1].oe,data[k*2].eo+data[k*2+1].ee));    Data[k].ee=max (data[k].ee,data[k*2].ee);    Data[k].ee=max (data[k].ee,data[k*2+1].ee);    Data[k].oo=max (-inf,max (data[k*2].oo+data[k*2+1].eo,data[k*2].oe+data[k*2+1].oo));    Data[k].oo=max (data[k].oo,data[k*2].oo);    Data[k].oo=max (data[k].oo,data[k*2+1].oo);    Data[k].eo=max (-inf,max (Data[k*2].ee+data[k*2+1].oo,data[k*2].eo+data[k*2+1].eo));    Data[k].eo=max (Data[k].eo,data[k*2].eo);    Data[k].eo=max (Data[k].eo,data[k*2+1].eo);    Data[k].oe=max (-inf,max (Data[k*2].oo+data[k*2+1].ee,data[k*2].oe+data[k*2+1].oe));    Data[k].oe=max (Data[k].oe,data[k*2].oe); Data[k].oe=max (Data[k].oe,data[k*2+1].oe);}   void build (int l,int r,int k) {int mid; Data[k].l=l;    Data[k].r=r;            if (l==r) {if (l%2==1) {data[k].oo=a[l];        Data[k].oe=data[k].eo=data[k].ee=-inf;            } else {data[k].ee=a[l];        Data[k].oe=data[k].eo=data[k].oo=-inf;    } return;    } mid= (L+r)/2;    Build (L,MID,K*2);    Build (mid+1,r,k*2+1); Pushup (k);}            void Updata (int n,int x,int k) {if (data[k].l==n && data[k].r==n) {if (data[k].l%2==1) {            Data[k].oo=x;        Data[k].eo=data[k].oe=data[k].ee=-inf;            } else {data[k].ee=x;        Data[k].eo=data[k].oe=data[k].oo=-inf;    } return;    } if (N&LT;=DATA[K*2].R) Updata (n,x,k*2);    else Updata (n,x,k*2+1); Pushup (k);}    Node search (int l,int r,int k) {node temp,t1,t2;    int mid;        if (data[k].l==l && data[k].r==r) {temp.ee=data[k].ee;        temp.oo=data[k].oo;        Temp.eo=data[k].eo; Temp.oe=data[k].oe;    return temp;    } mid= (DATA[K].L+DATA[K].R)/2;    if (R<=mid) return search (l,r,k*2);    else if (L>mid) return search (l,r,k*2+1);        else {t1=search (l,mid,k*2);        T2=search (mid+1,r,k*2+1);        Temp.ee=max (Max (Max (t1.ee+t2.oe,t1.eo+t2.ee), t1.ee), t2.ee);        Temp.eo=max (Max (Max (t1.eo+t2.eo,t1.ee+t2.oo), T1.eo), T2.eo);        Temp.oo=max (Max (Max (t1.oo+t2.eo,t1.oe+t2.oo), t1.oo), t2.oo);        Temp.oe=max (Max (Max (T1.oo+t2.ee,t1.oe+t2.oe), T1.oe), T2.oe);    return temp;    }}int Main () {int t,n,m,i,op,l,r;    Node Mark;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&m);        for (i=1;i<=n;i++) scanf ("%i64d", &a[i]);        Build (1,n,1);            while (m--) {scanf ("%d%d%d", &op,&l,&r);                if (op==0) {ans=-inf;                Mark=search (l,r,1);                Ans=max (ans,mark.oo);               Ans=max (ans,mark.ee); Ans=max (Ans,mark.eo);                Ans=max (Ans,mark.oe);            printf ("%i64d\n", ans);        } else Updata (l,r,1); }} return 0;}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

HDU 5316 segment tree interval maximum value problem