Magician
Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1613 Accepted Submission(s): 470
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 ?
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)
题目大意：有t组测试数据。每组有n，m。分别表示有n个精灵，m次询问。每个精灵有个能量值（可能为负值）。询问时，1 a,b表示将a位置的精灵赋值为b。0 a,b表示查询从a位置到b位置的精灵的美丽子序列的最大和。美丽子序列为下标呈现奇偶交替出现的子序列。
解题思路：这个题目很显然是线段树的题目。唯一比较难处理的是查询区间时需要是奇偶交替的子序列和。所以我们考虑四种情况，即 奇偶、偶奇、奇奇、偶偶。偶数下标开始奇数下标结尾的最大值可以由 左儿子的偶偶+右儿子的奇奇 得到，也可以由左儿子的偶奇+右儿子偶奇得到。也可以是直接由左儿子或者右儿子的偶奇得到的。类推其他三种情况。
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+20;
const int INF=0x3f3f3f3f;
typedef long long INT;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
#define max(a,b) ((a)>(b)?(a):(b))
INT a[maxn];
int n;
struct SegTree{
INT oj,jo,jj,oo;
void change(){
jo=jj=oo=oj=-INF;
}
}seg[maxn*4];
void PushUpDif(SegTree &ret,SegTree <ree,SegTree &rtree){
INT jjv1,jjv2,jov1,jov2,oov1,oov2,ojv1,ojv2;
jjv1=max(ltree.jj,rtree.jj);//从左右子树得到jj的情况
jjv2=max(ltree.jj+rtree.oj,ltree.jo+rtree.jj);//需要合并左子树jj右子树oj和左子树jo右子树jj的情况
ret.jj=max(jjv1,jjv2);//取最值
//下面三种同理可得
ojv1=max(ltree.oj,rtree.oj);
ojv2=max(ltree.oo+rtree.jj,ltree.oj+rtree.oj);
ret.oj=max(ojv1,ojv2);
jov1=max(ltree.jo,rtree.jo);
jov2=max(ltree.jj+rtree.oo,ltree.jo+rtree.jo);
ret.jo=max(jov1,jov2);
oov1=max(ltree.oo,rtree.oo);
oov2=max(ltree.oj+rtree.oo,ltree.oo+rtree.jo);
ret.oo=max(oov1,oov2);
} void build(int rt,int L,int R){
seg[rt].jo=seg[rt].oj=seg[rt].jj=seg[rt].oo=-INF;
if(L==R){
if(L%2==0){
seg[rt].oo=a[L];
}else{
seg[rt].jj=a[L];
}
return ;
}
build(lson);
build(rson);
PushUpDif(seg[rt],seg[rt*2],seg[rt*2+1]);
} void update(int rt,int L,int R,INT pos,INT val){
if(L==R){
if(L%2==0){
seg[rt].oo=val;
}else{
seg[rt].jj=val;
}
return ;
}
if(pos<=mid){
update(lson,pos,val);
}else{
update(rson,pos,val);
}
PushUpDif(seg[rt],seg[rt*2],seg[rt*2+1]);
} SegTree query(int rt,int L,int R,INT l_ran,INT r_ran){
if(l_ran<=L&&R<=r_ran){
return seg[rt];
}
SegTree ret,lret,rret;
ret.change(),lret.change(),rret.change();
if(l_ran<=mid){
lret=query(lson,l_ran,r_ran);
}
if(r_ran>mid){
rret=query(rson,l_ran,r_ran);
}
PushUpDif(ret,lret,rret);
return ret;
}
int main(){
int t,m,typ;
INT aa,bb;
SegTree ans;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
}
build(1,1,n);
for(int i=0;i<m;i++){
scanf("%d%lld%lld",&typ,&aa,&bb);
if(typ==1){
update(1,1,n,aa,bb);
}else{
ans=query(1,1,n,aa,bb);
INT ans1=max(ans.jj,ans.oo),ans2=max(ans.jo,ans.oj);
printf("%lld\n",max(ans1,ans2));
}
}
}
return 0;
}
/*
20
10 6
1 2 3 4 5 5 6 -4 5 10 0 1 5
0 4 7
0 4 8
1 1 -1
0 1 3
0 1 4 5 3
-5 7 -3 4 5 0 1 3
1 1 0
0 1 3 */