You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| ≤ 15007 , 1 ≤ N ≤ 50000 ). A query is defined as follows:
Query(x,y) = Max { a[i]+a[i+1]+...+a[j] ; x ≤ i ≤ j ≤ y }.
Given M queries, your program must output the results of these queries.

Input
The first line of the input file contains the integer N.
In the second line, N numbers follow.
The third line contains the integer M.
M lines follow, where line i contains 2 numbers xi and yi.
Output
Your program should output the results of the M queries, one query per line.

Sample Input
3
-1 2 3
1
1 2
Sample Output
2

題意：

裸題，沒有更新，只有查詢。

思路：

區(qū)間中維護(hù)一下值：

從左端點(diǎn)開始的連續(xù)的最大值lm

從右端點(diǎn)開始的連續(xù)最大值rm

區(qū)間的和sum。

區(qū)間中的連續(xù)最大值num。

那么更新操作為：

lm=max（左兒子的lm，左兒子的sum+右兒子的lm）

rm=max（右兒子的rm，右兒子的sum+左兒子的rm）

num=max（左兒子的num，右兒子的num，左兒子的rm+右兒子的lm）

詢問操作也返回區(qū)間，對(duì)區(qū)間進(jìn)行合并處理操作。

細(xì)節(jié)見代碼：

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl using namespace std; typedef long long ll; 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;} inline void getInt(int* p); const int maxn = 50000+7; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ struct node {int l;int r;ll num;ll lm;ll sum;ll rm; }segment_tree[maxn<<2]; int n; void pushup(int rt) {segment_tree[rt].sum=segment_tree[rt<<1].sum+segment_tree[rt<<1|1].sum;segment_tree[rt].lm=max(segment_tree[rt<<1].lm,segment_tree[rt<<1].sum+segment_tree[rt<<1|1].lm);segment_tree[rt].rm=max(segment_tree[rt<<1|1].rm,segment_tree[rt<<1|1].sum+segment_tree[rt<<1].rm);segment_tree[rt].num=max(segment_tree[rt<<1].num,segment_tree[rt<<1|1].num);segment_tree[rt].num=max(segment_tree[rt].num,segment_tree[rt<<1].rm+segment_tree[rt<<1|1].lm); }void build(int rt,int l,int r) {segment_tree[rt].l=l;segment_tree[rt].r=r;if(l==r){scanf("%lld",&segment_tree[rt].num);segment_tree[rt].lm=segment_tree[rt].rm=segment_tree[rt].num;segment_tree[rt].sum=segment_tree[rt].num;return ;}int mid=(l+r)>>1;build(rt<<1,l,mid);build(rt<<1|1,mid+1,r);pushup(rt); }node ask(int rt,int l,int r) {if(segment_tree[rt].l==l&&segment_tree[rt].r==r){return segment_tree[rt];}int mid=(segment_tree[rt].r+segment_tree[rt].l)>>1;if(l>mid){return ask(rt<<1|1,l,r);}else if(r<=mid){return ask(rt<<1,l,r);}else{node res1=ask(rt<<1,l,mid);node res2=ask(rt<<1|1,mid+1,r);node res;res.sum=res1.sum+res2.sum;res.lm=max(res1.lm,res1.sum+res2.lm);res.rm=max(res2.rm,res2.sum+res1.rm);res.num=max(res1.num,res2.num);res.num=max(res.num,res1.rm+res2.lm);return res;} }int main() {//freopen("D:\\code\\text\\input.txt","r",stdin);//freopen("D:\\code\\text\\output.txt","w",stdout);scanf("%d",&n);build(1,1,n);int m;scanf("%d",&m);while(m--){int x,y;scanf("%d %d",&x,&y);printf("%lld\n",ask(1,x,y).num);}return 0; }inline void getInt(int* p) {char ch;do {ch = getchar();} while (ch == ' ' || ch == '\n');if (ch == '-') {*p = -(getchar() - '0');while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 - ch + '0';}}else {*p = ch - '0';while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 + ch - '0';}} }

轉(zhuǎn)載于:https://www.cnblogs.com/qieqiemin/p/11438371.html

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