标签：线段树

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5316

Magician

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 ?

1
1 1
1
0 1 1

1

题意：有n个数，两个操作，0操作，输出l到r ，所有奇偶交替 的子序列中，值的最大和。（自序列是可以不用连续的）。 1操作是把a位置的数改成b。

做法：维护区间内的 

jiou，  这个区间以奇数位开始，偶数位结束的 所有子序列中的最大和。

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;
}

hdu 5316 Magician 线段树

标签：线段树

原文地址：http://blog.csdn.net/u013532224/article/details/47109575