Problem B: 取石子

标签：clu   一个   问题   cst   while   char   problem   queue   include   

• 转换成一个数在(0,X + Y)的加减问题
• 考虑一种使用线段树处理的方法， 维护前缀最大值， 前缀最小值， 前缀和， 然后查询的时候先询问右区间是否会同时碰到上下界， 会的话左区间无用直接递归右区间， 否则的话递归左区间， 然后右区间只会碰到上边界或者下边界， 分两种情况讨论即可

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define M 500010
#define ls now << 1
#define rs now << 1 | 1
#define lson l, mid, now << 1
#define rson mid + 1, r, now << 1 | 1
#define ll long long
#define mmp make_pair
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
int n, q, x, y, z;
ll sum[M << 2], minn[M << 2], maxx[M << 2];

void pushup(int now) {
    sum[now] = sum[ls] + sum[rs];
    minn[now] = min(minn[ls], sum[ls] + minn[rs]);
    maxx[now] = max(maxx[ls], sum[ls] + maxx[rs]);
}

void modify(int l, int r, int now, int pl, int v) {
    if(l > pl || r < pl) return;
    if(l == r) {
        sum[now] = v;
        minn[now] = min(v, 0);
        maxx[now] = max(v, 0);
        return;
    }
    int mid = (l + r) >> 1;
    modify(lson, pl, v), modify(rson, pl, v);
    pushup(now);
}
ll s;

ll merge(ll v, int now) {
    if(v + maxx[now] > s) return s - maxx[now] + sum[now];
    if(v + minn[now] < 0) return sum[now] - minn[now];
    return v + sum[now];
}

ll query(int l, int r, int now, ll v) {
    if(l == r) return max(min(s, sum[now] + v), 0ll);
    int mid = (l + r) >> 1;
    if(maxx[rs] - minn[rs] > s) return query(rson, 0);
    else return merge(query(lson, v), rs);
}

int main() {
//  freopen("stone1.in", "r", stdin);
    n = read(), q = read();
    x = read(), y = read();
    for(int i = 1; i <= n; i++) {
        z = read();
        if(i % 2 == 0) z = -z;
        modify(1, n, 1, i, z);
    }
    while(q--) {
        int op = read();
        if(op == 1) x = read();
        else if(op == 2) y = read();
        else {
            int pl = read(), v = read();
            if(pl % 2 == 0) v = -v;
            modify(1, n, 1, pl, v);
        }
        s = x + y;
        cout << query(1, n, 1, x) << "\n";
    }
    return 0;
}

Problem B: 取石子

标签：clu   一个   问题   cst   while   char   problem   queue   include   

原文地址：https://www.cnblogs.com/luoyibujue/p/10713334.html