hdu 5316 Magician 线段树

链接：http://acm.hdu.edu.cn/showproblem.php?

pid=5316

Magician

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 190    Accepted Submission(s): 54

Problem Description

Fantasy magicians usually gain their ability through one of three usual methods: possessing it as an innate talent, gaining it through study and practice, or receiving it from another being, often a god, spirit, or demon of some sort. Some wizards are depicted as having a special gift which sets them apart from the vast majority of characters in fantasy worlds who are unable to learn magic.

Magicians, sorcerers, wizards, magi, and practitioners of magic by other titles have appeared in myths, folktales, and literature throughout recorded history, with fantasy works drawing from this background.

In medieval chivalric romance, the wizard often appears as a wise old man and acts as a mentor, with Merlin from the King Arthur stories representing a prime example. Other magicians can appear as villains, hostile to the hero.



Mr. Zstu is a magician, he has many elves like dobby, each of which has a magic power (maybe negative). One day, Mr. Zstu want to test his ability of doing some magic. He made the elves stand in a straight line, from position 1 to position n, and he used two kinds of magic, Change magic and Query Magic, the first is to change an elf’s power, the second is get the maximum sum of beautiful subsequence of a given interval. A beautiful subsequence is a subsequence that all the adjacent pairs of elves in the sequence have a different parity of position. Can you do the same thing as Mr. Zstu ?

 

Input

The first line is an integer T represent the number of test cases.
Each of the test case begins with two integers n, m represent the number of elves and the number of time that Mr. Zstu used his magic.
(n,m <= 100000)
The next line has n integers represent elves’ magic power, magic power is between -1000000000 and 1000000000.
Followed m lines, each line has three integers like 
type a b describe a magic.
If type equals 0, you should output the maximum sum of beautiful subsequence of interval [a,b].(1 <= a <= b <= n)
If type equals 1, you should change the magic power of the elf at position a to b.(1 <= a <= n, 1 <= b <= 1e9)

 

Output

For each 0 type query, output the corresponding answer.

 

Sample Input

1
1 1
1
0 1 1

 

Sample Output

1

 

题意：有n个数。两个操作，0操作，输出l到r 。全部奇偶交替 的子序列中。值的最大和。

（自序列是能够不用连续的）。 1操作是把a位置的数改成b。

做法：维护区间内的 

jiou，  这个区间以奇数位開始，偶数位结束的 全部子序列中的最大和。

ouji，jiji。ouou 三个数同理。

#include <cstdio>
#include <algorithm>
#include <stdio.h>
#include <string.h>
using namespace std;
 
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define LL __int64
#define inf -100000000000000000

const int maxn = 111111;
struct point 
{
    __int64 jiji,ouou,jiou,ouji;
};


point pp[maxn<<2]; 
int jiou[maxn<<2];


point big(point a,point b)
{
    point c;
    c.jiji=max(a.jiji,b.jiji);
    c.jiji=max(c.jiji,a.jiji+b.ouji);
    c.jiji=max(c.jiji,a.jiou+b.jiji);
    c.jiji=max(c.jiji,inf);


    c.jiou=max(a.jiou,b.jiou);
    c.jiou=max(c.jiou,a.jiji+b.ouou);
    c.jiou=max(c.jiou,a.jiou+b.jiou);
    c.jiou=max(c.jiou,inf);

    c.ouji=max(a.ouji,b.ouji);
    c.ouji=max(c.ouji,a.ouji+b.ouji);
    c.ouji=max(c.ouji,a.ouou+b.jiji);
    c.ouji=max(c.ouji,inf);


    c.ouou=max(a.ouou,b.ouou);
    c.ouou=max(c.ouou,a.ouji+b.ouou);
    c.ouou=max(c.ouou,a.ouou+b.jiou);
    c.ouou=max(c.ouou,inf);
    return c;
}


void PushUp(int rt) {
    pp[rt]=big(pp[rt<<1],pp[rt<<1|1]);
}

int kk=1;
void build(int l,int r,int rt) {
    if (l == r) {
        if(kk&1)
        {
            jiou[rt]=1;
            scanf("%I64d",&pp[rt].jiji);
            pp[rt].ouou=inf;
            pp[rt].ouji=inf;
            pp[rt].jiou=inf;
        }
        else
        {
            jiou[rt]=0;
            scanf("%I64d",&pp[rt].ouou);
            pp[rt].jiou=pp[rt].jiji=pp[rt].ouji=inf;
        }
        kk++;
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}
void update(int L,int c,int l,int r,int rt) {
    if (l==r&&l==L) { 
        if(jiou[rt]==1)
            pp[rt].jiji=c;
        if(jiou[rt]==0)
            pp[rt].ouou=c;
        return ;
    }
    int m = (l + r) >> 1;
    if (L <= m) update(L , c , lson);
    else
        update(L  , c , rson);
    PushUp(rt);
}
point query(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) { 
        return pp[rt];
    }

    int m = (l + r) >> 1;

    if(L<=m&&m<R)  return big( query(L , R , lson),query(L , R , rson));
    if(L<=m) return query(L , R , lson);
    if(m<R) return query(L , R , rson);
      
}
int main() {
    int t;
    int n,q;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&q);
        kk=1;
        if(n)
        build(1 , n , 1);
        while(q--)
        {
            int op; 
            int a,b;
            scanf("%d",&op);
            scanf("%d%d",&a,&b);
            if(n==0)
            {
                if(op==0)
                    puts("0"); 
                continue;
            }
            
            if(op==0)
            {
                if(a>b)
                    swap(a,b);
                point ppp=query(a,b,1,n,1);
                printf("%I64d\n",max(ppp.jiji,max(ppp.jiou,max(ppp.ouji,ppp.ouou))));
            }
            else
                update(a , b , 1 , n , 1);
        } 
    } 
    return 0;
}