Luogu2161 [SHOI2009]会场预约-线段树

Solution

线段树维护 sum 表示区间内预约个数， L 表示区间最左边的预约， R 表示区间最右边的预约。

$pushup$ 就是这样 ： 

void up(int nd) {
        sum[nd] = sum[lson] + sum[rson] - (R[lson] == L[rson] && R[lson]);
        L[nd] = L[lson]; R[nd] = R[rson];
}

每次查询答案类似于$pushup$。 

考虑把旧预约删去 ：

若该次新预约 开始时间为$l$， 结束时间为$r$，是第$i$ 个预约， 则将 与该区间在右边 和 在左边相交的 旧预约的区间都用0覆盖。

然后再把新预约的区间都用 $i$覆盖。

$pushdown$ 和 $laz$ 数组也有些细节

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define rd read()
using namespace std;

const int N = 2e5 + 5;
const int end = 1e5;

struct met {
    int L, R;
}in[N];

int read() {
    int X = 0, p = 1; char c = getchar();
    for (; c > '9' || c < '0'; c = getchar())
        if (c == '-') p = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        X = X * 10 + c - '0';
    return X * p;
}

namespace SegT {
    int sum[N << 2], L[N << 2], R[N << 2], laz[N << 2];
    struct node {
        int sum, L, R;
    };
#define lson nd << 1
#define rson nd << 1 | 1
#define mid ((l + r) >> 1)
    void up(int nd) {
        sum[nd] = sum[lson] + sum[rson] - (R[lson] == L[rson] && R[lson]);
        L[nd] = L[lson]; R[nd] = R[rson];
    }

    void make(int nd, int x) {
        sum[nd] = x ? 1 : 0; L[nd] = R[nd] = laz[nd] = x;
    }

    void down(int nd) {
        if (~laz[nd]) {
            make(lson, laz[nd]); make(rson, laz[nd]);
            laz[nd] = -1;
        }
    }

    void modify(int cl, int cr, int c, int l, int r, int nd) {
        if (cl <= l && r <= cr) {
            make(nd, c); return;
        }
        down(nd);
        if (mid >= cl)
            modify(cl, cr, c, l, mid, lson);
        if (mid < cr)
            modify(cl, cr, c, mid + 1, r, rson);
        up(nd);
    }

    node query(int cl, int cr ,int l, int r, int nd) {
        if (cl <= l && r <= cr) {
            node tmp;
            tmp.sum = sum[nd]; tmp.L = L[nd]; tmp.R = R[nd];
            return tmp;
        }
        node tmp, ls, rs;
        down(nd);
        if (mid >= cl && !(mid < cr))
            return query(cl, cr, l, mid, lson);
        else if (!(mid >= cl) && mid < cr)
            return query(cl, cr, mid + 1, r, rson);
        ls = query(cl, cr, l, mid, lson);
        rs = query(cl, cr, mid + 1, r, rson);
        tmp.sum = ls.sum + rs.sum - (ls.R == rs.L && ls.R);
        tmp.L = ls.L; tmp.R = rs.R;
        return tmp;
    }
}using namespace SegT;

int main()
{
    int n = rd;
    memset(laz, -1, sizeof(laz));
    for (int i = 1; i <= n; ++i) {
        char ch = getchar();
        while (ch > 'Z' || ch < 'A') ch = getchar();
        if (ch == 'A') {
            int l = rd, r = rd;
            in[i].L = l; in[i].R = r;
            node ans = query(l, r, 1, end, 1);
            printf("%d
", ans.sum);
            if (ans.L) modify(in[ans.L].L, in[ans.L].R, 0, 1, end, 1);
            if (ans.R) modify(in[ans.R].L, in[ans.R].R, 0, 1, end, 1);
            modify(l, r, i, 1, end, 1);
        }
        else {
            node ans = query(1, end, 1, end, 1);
            printf("%d
", ans.sum);
        }
    }
}