hdu5316 Magician (线段树+单点更新+区间查询+区间合并)

Magician
Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3084 Accepted Submission(s): 838

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：给你n个数，有两种操作：
1ab 将下标为a的数变为b
0lr 求[l,r]内的一组数，这组数相邻的数的下标的奇偶性不同，求这组数的最大和.

思路：满足条件的一组数的首尾可能有四种情况：奇奇、奇偶、偶偶、偶奇，所以说，我们在线段树中加上这四种情况来维护，在区间查询和单点查询的时候和平常的一样，加上了一个区间合并。

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<map>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0x7fffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define eps 1e-8using namespace std;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}//headstruct s{    ll lc,rc,jj,jo,oo,oj;}tree[MAXN],kk;ll num[MAXN];s findmax(int i,s A,s B){    s k;ll a,b,c,d;    a=max(A.jj,B.jj);a=max(a,A.jj+B.oj);a=max(a,A.jo+B.jj);    b=max(A.jo,B.jo);b=max(b,A.jo+B.jo);b=max(b,A.jj+B.oo);    c=max(A.oo,B.oo);c=max(c,A.oo+B.jo);c=max(c,A.oj+B.oo);    d=max(A.oj,B.oj);d=max(d,A.oj+B.oj);d=max(d,A.oo+B.jj);    tree[i].jj=a;tree[i].jo=b;tree[i].oo=c;tree[i].oj=d;    return k;}s getmax(s A,s B){    s k;ll a,b,c,d;    a=max(A.jj,B.jj);a=max(a,A.jj+B.oj);a=max(a,A.jo+B.jj);    b=max(A.jo,B.jo);b=max(b,A.jo+B.jo);b=max(b,A.jj+B.oo);    c=max(A.oo,B.oo);c=max(c,A.oo+B.jo);c=max(c,A.oj+B.oo);    d=max(A.oj,B.oj);d=max(d,A.oj+B.oj);d=max(d,A.oo+B.jj);    k.jj=a;k.jo=b;k.oo=c;k.oj=d;    return k;}void build(int i,ll l,ll r){    tree[i].lc=l,tree[i].rc=r;    if(l==r)    {        if(l%2)        {            tree[i].jj=num[l];tree[i].jo=-INF;            tree[i].oo=-INF;tree[i].oj=-INF;        }        else        {            tree[i].jj=-INF;tree[i].jo=-INF;            tree[i].oo=num[l];tree[i].oj=-INF;        }        return ;    }    else    {        int mid=(l+r)/2;        build(i*2,l,mid);build(i*2+1,mid+1,r);        findmax(i,tree[i*2],tree[i*2+1]);    }}void update(ll aa,ll b,int i){    if(tree[i].lc==tree[i].rc&&tree[i].lc==aa)    {        if(aa%2) tree[i].jj=b;        else tree[i].oo=b;        return;    }    ll mid=(tree[i].lc+tree[i].rc)/2;    if(mid>=aa)        update(aa,b,i*2);    else        update(aa,b,i*2+1);    findmax(i,tree[i*2],tree[i*2+1]);}s query(int i,ll l,ll r){    if(tree[i].lc==l&&tree[i].rc==r)        return tree[i];    ll mid=(tree[i].lc+tree[i].rc)/2;    if(r<=mid)        kk=query(i*2,l,r);    else if(l>mid)        kk=query(i*2+1,l,r);    else        kk=getmax(query(i*2,l,mid),query(i*2+1,mid+1,r));    return kk;}int main(){    int t;scanf("%d",&t);    while(t--)    {        ll n,m;scanf("%I64d%I64d",&n,&m);        for(ll i=1;i<=n;i++) scanf("%I64d",&num[i]);        build(1,1,n);        while(m--)        {            int ch;ll b,c;scanf("%d%I64d%I64d",&ch,&b,&c);            if(ch==1)                update(b,c,1);            else if(ch==0)            {                s ans=query(1,b,c);                printf("%I64d\n",max(max(ans.jj,ans.jo),max(ans.oj,ans.oo)));            }        }    }    return 0;}