spoj GSS1.区间最大子段和(线段树)

区间最大子段和

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 $x_i$ and $y_i$ .

Output
Your program should output the results of the M queries, one query per line.

Example
Input:

3
-1 2 3
1
1 2

Output:

2

Solution

对于每一个每个区间，保存一个Node，记录这个区间的sum(总和)，maxsum(最大子段和)，lmax(最大前缀和)，rmax(最大后缀和)，有了这四个标记，就可以随意转移，为所欲为了。。。
这里我们把核心代码拉出来讲一讲

ans.sum = lo.sum + ro.sum;//sum直接相加
ans.maxsum = max(max(lo.maxsum, ro.maxsum), lo.rmax + ro.lmax);//左右两半的最大子段和，合并起来的最大子段和
ans.lmax = max(lo.lmax, lo.sum + ro.lmax);//左边的最大前缀和，左边整段+右边最大前缀
ans.rmax = max(ro.rmax, ro.sum + lo.rmax);//右边的最大后缀和，右边整段+左边最大后缀

Code

#include <cstdio>
#include <algorithm>
#define N 50010

using namespace std;

struct Node{
    int sum, maxsum, lmax, rmax;
}tr[N << 2];
int a[N];

void build(int l, int r, int id) {
    if (l == r) {
        tr[id].sum = tr[id].maxsum = tr[id].lmax = tr[id].rmax = a[l];
        return;
    }
    int mid = (l + r) >> 1, ls = id << 1, rs = id << 1 | 1;
    build(l, mid, ls);
    build(mid + 1, r, rs);
    tr[id].sum = tr[ls].sum + tr[rs].sum;
    tr[id].maxsum = max(max(tr[ls].maxsum, tr[rs].maxsum), tr[ls].rmax + tr[rs].lmax);
    tr[id].lmax = max(tr[ls].lmax, tr[ls].sum + tr[rs].lmax);
    tr[id].rmax = max(tr[rs].rmax, tr[rs].sum + tr[ls].rmax);
}

Node query(int l, int r, int ll, int rr, int id) {
    if (ll <= l && r <= rr) {
        return tr[id];
    }
    int mid = (l + r) >> 1, ls = id << 1, rs = id << 1 | 1;
    if (rr <= mid) return query(l, mid, ll, rr, ls);
    if (mid < ll) return query(mid + 1, r, ll, rr, rs);
    Node lo = query(l, mid, ll, rr, ls), ro = query(mid + 1, r, ll, rr, rs), ans;
    ans.sum = lo.sum + ro.sum;
    ans.maxsum = max(max(lo.maxsum, ro.maxsum), lo.rmax + ro.lmax);
    ans.lmax = max(lo.lmax, lo.sum + ro.lmax);
    ans.rmax = max(ro.rmax, ro.sum + lo.rmax);
    return ans;
}

int main() {
    int n, m;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    build(1, n, 1);
    scanf("%d", &m);
    for (int i = 1; i <= m; ++i) {
        int x, y;
        scanf("%d%d", &x, &y);
        printf("%d\n", query(1, n, x, y, 1).maxsum);
    }
    return 0;
}