2018.07.08 hdu5316 Magician（线段树）

Magician

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

Author

ZSTU

Source

2015 Multi-University Training Contest 3

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仍然是简单线段树，单点修改，要求从区间中选出一个非空子数列并保证相邻元素在原数组中下标奇偶性不同，求数列总和最大值 。

so" role="presentation" style="position: relative;">soso how" role="presentation" style="position: relative;">howhow to" role="presentation" style="position: relative;">toto solve" role="presentation" style="position: relative;">solvesolve this" role="presentation" style="position: relative;">thisthis easy" role="presentation" style="position: relative;">easyeasy problem?" role="presentation" style="position: relative;">problem?problem?

n≤50000" role="presentation" style="position: relative;">n≤50000n≤50000，我会分块！

n≤100000" role="presentation" style="position: relative;">n≤100000n≤100000+多组数据，线段树？

线段树！

对于每个节点，我们用ll,lr,rl,rr" role="presentation" style="position: relative;">ll,lr,rl,rrll,lr,rl,rr来记录区间内以奇（偶）下标作为左端点，奇（偶）下标作为右端点时区间中满足条件的最大和，对于每种情况稍（du）微(liu)讨论一下就行了。然而本蒟蒻忘记输出时要转成long" role="presentation" style="position: relative;">longlong long" role="presentation" style="position: relative;">longlong然后就gg" role="presentation" style="position: relative;">gggg了。

代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define inf 0xffffffffffff
#define N 100005
#define lc (p<<1)
#define rc (p<<1|1)
#define mid (T[p].l+T[p].r>>1)
using namespace std;
inline long long read(){
    long long ans=0,w=1;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')w=-1;
        ch=getchar();
    }
    while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-'0',ch=getchar();
    return ans*w;
}
long long n,a[N],m,t;
struct Node{long long l,r,ll,rr,lr,rl;}T[N<<2];
inline void pushup(long long p){
    T[p].ll=max(T[lc].ll,T[rc].ll);
    T[p].ll=max(T[lc].ll+T[rc].rl,T[p].ll);
    T[p].ll=max(T[lc].lr+T[rc].ll,T[p].ll);
    T[p].rr=max(T[lc].rr,T[rc].rr);
    T[p].rr=max(T[lc].rr+T[rc].lr,T[p].rr);
    T[p].rr=max(T[lc].rl+T[rc].rr,T[p].rr);
    T[p].lr=max(T[lc].lr,T[rc].lr);
    T[p].lr=max(T[lc].lr+T[rc].lr,T[p].lr);
    T[p].lr=max(T[lc].ll+T[rc].rr,T[p].lr);
    T[p].rl=max(T[lc].rl,T[rc].rl);
    T[p].rl=max(T[lc].rl+T[rc].rl,T[p].rl);
    T[p].rl=max(T[lc].rr+T[rc].ll,T[p].rl);
}
inline void build(long long p,long long l,long long r){
    T[p].l=l,T[p].r=r;
    if(l==r){
        T[p].ll=T[p].lr=T[p].rr=T[p].rl=-inf;
        if(l&1)T[p].ll=a[l];
        else T[p].rr=a[l];
        return;
    }
    build(lc,l,mid);
    build(rc,mid+1,r);
    pushup(p);
}
inline void update(long long p,long long k,long long v){
    if(T[p].l==T[p].r){
        if(T[p].l&1)T[p].ll=v;
        else T[p].rr=v;
        return;
    }
    if(k<=mid)update(lc,k,v);
    else update(rc,k,v);
    pushup(p);
}
inline Node query(long long p,long long ql,long long qr){
    if(ql<=T[p].l&&T[p].r<=qr)return T[p];
    if(qr<=mid)return query(lc,ql,qr);
    if(ql>mid)return query(rc,ql,qr);
    Node ans1=query(lc,ql,mid),ans2=query(rc,mid+1,qr),ans;
    ans.l=ans1.l,ans.r=ans2.r;
    ans.ll=max(ans1.ll,ans2.ll);
    ans.ll=max(ans1.ll+ans2.rl,ans.ll);
    ans.ll=max(ans1.lr+ans2.ll,ans.ll);
    ans.rr=max(ans1.rr,ans2.rr);
    ans.rr=max(ans1.rr+ans2.lr,ans.rr);
    ans.rr=max(ans1.rl+ans2.rr,ans.rr);
    ans.lr=max(ans1.lr,ans2.lr);
    ans.lr=max(ans1.lr+ans2.lr,ans.lr);
    ans.lr=max(ans1.ll+ans2.rr,ans.lr);
    ans.rl=max(ans1.rl,ans2.rl);
    ans.rl=max(ans1.rl+ans2.rl,ans.rl);
    ans.rl=max(ans1.rr+ans2.ll,ans.rl);
    return ans;
}
int main(){
    t=read();
    while(t--){
        n=read(),m=read();
        for(long long i=1;i<=n;++i)a[i]=read();
        build(1,1,n);
        while(m--){
            long long op=read(),l=read(),r=read();
            switch(op){
                case 0:{
                    Node q=query(1,l,r);
                    printf("%lld\n",max(max(q.ll,q.lr),max(q.rl,q.rr)));
                    break;
                }
                default:{
                    update(1,l,r);
                    break;
                }
            }
        }
    }
    return 0;
}