P3071 [USACO13JAN]座位Seating

P3071 [USACO13JAN]座位Seating

题目链接

思路:

一开始把题给读错了浪费了好多时间呜呜呜。
因为第二个撤离操作是区间修改，所以我们可以想到用线段树来做。对于第一个操作，我们只需要维护suml,sumr,sum分别表示当前结点左端连续有多少个空位、右端连续有多少个空位、以及最长连续空位为多少就行了。因为每次安排作为可能会跨过mid，所以我们还需要数组来维护一下信息。
注意一下代码的细节吧：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5e5 + 5;
int n, m;
int c[N];
char s[2] ;
int sum[N << 2], lazy[N << 2], suml[N << 2], sumr[N << 2];
void build(int o, int l, int r) {
    sum[o] = suml[o] = sumr[o] = r - l + 1;
    if(l == r) return ;
    int mid = (r + l) >> 1;
    build(o << 1, l, mid);
    build(o << 1|1, mid + 1, r) ;
}
void pushdown(int o, int l, int r) {
    if(lazy[o] != 0) {
        int mid = (l + r) >> 1 ;
        suml[o << 1] = sumr[o << 1] = sum[o << 1] = (lazy[o] > 0) * (mid - l + 1);
        suml[o << 1|1] = sumr[o << 1|1] = sum[o << 1|1] = (lazy[o] > 0) * (r - mid);
        lazy[o << 1|1] = lazy[o << 1] = lazy[o] ;
        lazy[o] = 0;
    }
}
void push_up(int o, int l, int r) {
    int mid = (r + l) >> 1;
    sum[o] = max(max(sum[o << 1], sum[o << 1|1]), sumr[o << 1] + suml[o << 1|1]) ;
    suml[o] = suml[o << 1] + (suml[o << 1] == (mid - l + 1) ? suml[o << 1|1] : 0) ;
    sumr[o] = sumr[o << 1|1] + (sumr[o << 1|1] == (r - mid) ? sumr[o << 1] : 0) ;
}
void update(int o, int l, int r, int L, int R, int sign) {
    if(L <= l && r <= R) {
        lazy[o] = sign ;
        suml[o] = sumr[o] = sum[o] = (sign > 0) * (r - l + 1) ;
        return ;
    }
    pushdown(o, l, r) ;
    int mid = (l + r) >> 1;
    if(L <= mid) update(o << 1, l, mid, L, R, sign) ;
    if(R > mid) update(o << 1|1, mid + 1, r, L, R, sign) ;
    push_up(o, l, r) ;
}
int ask(int o, int l, int r, int x) {
    int mid = (l + r) >> 1;
    pushdown(o, l, r) ;
    if(sum[o << 1] >= x) return ask(o << 1, l, mid, x) ;
    else if(sumr[o << 1] + suml[o << 1|1] >= x) return mid - sumr[o << 1] + 1;
    else return ask(o << 1|1, mid + 1,r ,x);
}
int main() {
    cin >> n >> m;
    int ans = 0;
    build(1, 1, n) ;
    for(int i = 1, a, b; i <= m; i++) {
        scanf("%s", s);
        if(s[0] == 'A') {
            scanf("%d", &a);
            if(sum[1] < a) ans++;
            else {
                int pos = ask(1, 1, n, a) ;
                update(1, 1, n, pos, pos + a -1, -1) ;
            }
        } else {
            scanf("%d%d", &a, &b);
            update(1, 1, n, a, b, 1) ;
        }
    }
    cout << ans;
    return 0;
}

标签：P3071,空位,Seating,return,int,sum,mid,suml,USACO13JAN
来源： https://www.cnblogs.com/heyuhhh/p/10891279.html