有n个水塔，初始每个水塔有a[i]的水，每个水塔一次最多拿b[i]的水，现从1~n依次在水塔中取水，没取完的水全部流入下一个水塔，求最终能取多少水

题目

思路：

假设有两个水塔1和2，分类讨论：

1、当a1 > b1时，2中剩下的水是a2 - b2 + a1 - b1

2、当a1 < b1时，1中的水不会流到2中，2中剩下的水是a2 - b2

即最大（a - b) 的后缀和

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 5e5 + 5, maxm = 2e3 + 5;
int a[maxn], b[maxn], c[maxn], d[maxn], t[maxn << 2], suf[maxn];
int n;
int lazy[maxn << 2];
void push_up(int p){
	t[p] = max(t[lson], t[rson]);
}
void push_down(int p, int l, int r){
	lazy[lson] += lazy[p];
	lazy[rson] += lazy[p];
	int mid = (l + r) >> 1;
	t[lson] += lazy[p];
	t[rson] += lazy[p];
	lazy[p] = 0;
}
void build(int p, int l, int r){
	if(l == r){
		t[p] = suf[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(lson, l, mid);
	build(rson, mid + 1, r);
	push_up(p);
}
void update(int p, int l, int r, int L, int R, int val){
	if(L <= l && R >= r){
		t[p] += val;
		lazy[p] += val;
		return;
	}
	int mid = (l + r) >> 1;
	push_down(p, l, r);
	if(L <= mid) update(lson, l, mid, L, R, val);
	if(R >= mid + 1) update(rson, mid + 1, r, L, R, val);
	push_up(p);
}
void solve()
{
	cin >> n;
	int q;
	cin >> q;
	int suma = 0;
	for(int i = 1; i <= n; i++){
		cin >> a[i];
		suma += a[i];
	}
	for(int i = 1; i <= n; i++){
		cin >> b[i];
		d[i] = a[i] - b[i];
	}
	for(int i = 1; i < n; i++){
		cin >> c[i];
	}
	suf[n + 1] = 0;
	for(int i = n; i >= 1; i--){
		suf[i] = suf[i + 1] + d[i];
	}
	build(1, 1, n);
	while(q--){
		int p, x, y, z;
		cin >> p >> x >> y >> z;
		suma += x - a[p];
		int dif = (x - y) - d[p];
		a[p] = x;
		b[p] = y;
		d[p] = x - y;
		update(1, 1, n, 1, p, dif);
		int res = suma - max(0LL, t[1]);
		cout << res << '\n';
	}
}

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	int T = 1;
	// cin >> T;
	while (T--)
	{
		solve();
	}
}