[POJ2750]Potted Flower

Description

The little cat takes over the management of a new park. There is a large circular statue in the center of the park, surrounded by N pots of flowers. Each potted flower will be assigned to an integer number (possibly negative) denoting how attractive it is. See the following graph as an example:(Positions of potted flowers are assigned to index numbers in the range of 1 ... N. The i-th pot and the (i + 1)-th pot are consecutive for any given i (1 <= i < N), and 1st pot is next to N-th pot in addition.)



The board chairman informed the little cat to construct "ONE arc-style cane-chair" for tourists having a rest, and the sum of attractive values of the flowers beside the cane-chair should be as largeas possible. You should notice that a cane-chair cannot be a total circle, so the number of flowersbeside the cane-chair may be 1, 2, ..., N - 1, but cannot be N. In the above example, if we construct a cane-chair in the position of that red-dashed-arc, we will have the sum of 3+(-2)+1+2=4, which is the largest among all possible constructions.Unluckily, some booted cats always make trouble for the little cat, by changing some potted flowers to others. The intelligence agency of little cat hascaught up all the M instruments of booted cats' action. Each instrument is in the form of "A B", which means changing the A-th potted flowered with a new one whose attractive value equals to B. You have to report the new "maximal sum" after each instruction.

给定一个环形序列，进行在线操作，每次修改一个元素，输出环上的最大连续子段的和。

Input

There will be a single test data in the input. You are given an integer N (4 <= N <= 100000) in the

first input line.The second line contains N integers, which are the initial attractive value of eachpotted flower. The i-th number is for the potted flower on the i-th position.A single integer M (4

<= M <= 100000) in the third input line, and the following M lines each contains an instruction "A B

" in the form described above.Restriction: All the attractive values are within [-1000, 1000]. We gu

arantee the maximal sum will be always a positive integer.

Output

For each instruction, output a single line with the maximum sum of attractive values for the optimumcane-chair.

Sample Input

5

3 -2 1 2 -5

4

2 -2

5 -5

2 -4

5 -1

Sample Output

4

4

3

5

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破环成链，记录子序列最大值和最小值，以及区间和。

当环上所有的数都是正数时，答案为 区间和-子序列最小值

否则，答案为 max{区间和-子序列最小值，区间最大值}

有类似的题目，参考最大连续子数列和

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
	int x=0,f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x>=10)     print(x/10);
	putchar(x%10+'0');
}
const int N=1e5;
struct Segment{
	#define ls (p<<1)
	#define rs (p<<1|1)
	struct AC{
		int Maxl,Maxr,Max;//Max为子序列最大值，Maxl与Maxr是为了更新Max而产生，分别是从左边开始的序列最大值和从右边开始的序列最大值
		int Minl,Minr,Min;//与Max类似
		int sum;//区间和
		void init(int x){Maxl=Maxr=Max=Minl=Minr=Min=sum=x;}
	}tree[N*4+10];
	friend AC operator +(const AC &x,const AC &y){//重定义+后的更新
		AC z; z.init(0);
		z.Max=max(max(x.Max,y.Max),x.Maxr+y.Maxl);
		z.Maxl=max(x.Maxl,x.sum+y.Maxl);
		z.Maxr=max(y.Maxr,y.sum+x.Maxr);
		z.Min=min(min(x.Min,y.Min),x.Minr+y.Minl);
		z.Minl=min(x.Minl,x.sum+y.Minl);
		z.Minr=min(y.Minr,y.sum+x.Minr);
		z.sum=x.sum+y.sum;
		return z;
	}
	void build(int p,int l,int r){
		if (l==r){
			tree[p].init(read());
			return;
		}
		int mid=(l+r)>>1;
		build(ls,l,mid),build(rs,mid+1,r);
		tree[p]=tree[ls]+tree[rs];
	}
	void change(int p,int l,int r,int x,int t){
		if (l==r){
			tree[p].init(t);
			return;
		}
		int mid=(l+r)>>1;
		if (x<=mid)	change(ls,l,mid,x,t);
		if (x>mid)	change(rs,mid+1,r,x,t);
		tree[p]=tree[ls]+tree[rs];
	}
	void write(){
		int Ans=tree[1].sum-tree[1].Min;
		if (tree[1].sum!=tree[1].Max)	Ans=max(Ans,tree[1].Max);
		printf("%d\n",Ans);
	}
}T;
int main(){
	int n=read();
	T.build(1,1,n);
	int m=read();
	for (int i=1;i<=m;i++){
		int x=read(),y=read();
		T.change(1,1,n,x,y);
		T.write();
	}
	return 0;
}