2020ICPC小米網路賽 C.Data Structure Problem

題目連結
題目描述
Bobo has two sequences $a_1,a_2,\dots ,a_n$ and $b_1,b_2,\dots ,b_n$. He would like to perform the following operations:

• $\text{1 }xy$, change the value of $a_x$ to $y$.
• $\text{2 }xy$, change the value of $b_x$ to $y$.
• $\text{3 }x$, find the value of $c_x$, where $c_0=0$, $c_i=\max (c_{i-1}+b_i,a_i)\text{ for }1\le i\le x$.

輸入描述:
The input consists of several test cases terminated by end-of-file. For each test case:

The first line contains two integers $n$ and $m$ , which are the length of the two sequences and the number of operations. The second line contains $n$ integers $a_1,a_2,\dots ,a_n$. The third line contains $n$ integers $b_1,b_2,\dots ,b_n$. Each of the last m lines contains a query.

• $1\le n,m\le 2\times 10^5$

• $-10^9\le a_i,b_i,y\le 10^9$

• $1\le x\le n$

• The sum of $n$ and the sum of $m$ do not exceed $2\times 10^6$.

輸出描述:
For each query of $\text{3 }x$, output an integer denoting the value of $c_x$.
示例1
輸入

4 9
1 2 3 3
-1 2 3 3
3 1
3 2
3 3
3 4
2 2 -4
3 1
3 2
3 3
3 4

輸出

1
3
6
9
1
2
5
8

定義 $ans(l,r)=c_r$ ，$c_{l-1}=0$ ， $c_i=\max (c_{i-1}+b_i,a_i)\text{ for }l\le i\le r$ ；$sum(l,r)=\sum _{i=l}^r{b}_i$ 。
則對於兩個區間 $[l,x]$ 和 $[x+1,r]$ ，滿足 $ans(l,r)=\max (ans(l,x)+sum(x+1,r),ans(x+1,r))$ 。
用反證法很好證明。用線段樹維護即可。

#include<bits/stdc++.h>

#define ll long long
#define fi first
#define se second
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 2e5 + 10;
const ll INF = 4e18;

struct Seg_Tree {
    struct {
        int l, r;
        ll sum, ans;
    } t[N * 4];

#define update(x) {                                                          \
        t[x].sum = t[x << 1].sum + t[x << 1 | 1].sum;                        \
        t[x].ans = max(t[x << 1].ans + t[x << 1 | 1].sum, t[x << 1 | 1].ans);\
    }                                                                        \


    void build(int p, int l, int r, const int *a, const int *b) {
        t[p].l = l, t[p].r = r;
        if (l == r) {
            t[p].ans = a[l];
            t[p].sum = b[l];
            return;
        }
        int mid = (l + r) >> 1;
        build(p << 1, l, mid, a, b);
        build(p << 1 | 1, mid + 1, r, a, b);
        update(p)
    }

    void changea(int p, int x, int v) {
        if (t[p].l == t[p].r) {
            t[p].ans = v;
            return;
        }
        int mid = (t[p].l + t[p].r) >> 1;
        if (x <= mid)changea(p << 1, x, v);
        else changea(p << 1 | 1, x, v);
        update(p)
    }

    void changeb(int p, int x, int v) {
        if (t[p].l == t[p].r) {
            t[p].sum = v;
            return;
        }
        int mid = (t[p].l + t[p].r) >> 1;
        if (x <= mid)changeb(p << 1, x, v);
        else changeb(p << 1 | 1, x, v);
        update(p)
    }

    pair<ll, ll> ask(int p, int x) {
        if (t[p].l > x)return {0, -INF};
        if (t[p].r <= x)return {t[p].sum, t[p].ans};
        pair<ll, ll> val = ask(p << 1, x);
        pair<ll, ll> res = ask(p << 1 | 1, x);
        val.fi += res.fi, val.se = max(val.se + res.fi, res.se);
        return val;
    }
} S;

int a[N], b[N], n, m;

int main() {
    while (~scanf("%d%d", &n, &m)) {
        for (int i = 2; i <= n + 1; ++i)a[i] = qr();
        for (int i = 2; i <= n + 1; ++i)b[i] = qr();
        S.build(1, 1, n + 1, a, b);
        for (int i = 1; i <= m; ++i)
            switch (qr()) {
                case 1: {
                    int x = qr(), y = qr();
                    S.changea(1, x + 1, y);
                    break;
                }
                case 2: {
                    int x = qr(), y = qr();
                    S.changeb(1, x + 1, y);
                    break;
                }
                default:
                    printf("%lld\n", S.ask(1, qr() + 1).se);
            }
    }
    return 0;
}