poj 1823 Hotel

http://poj.org/problem?id=1823

　　这是一道线段树的题，很久以前就听说它很经典！这是【成段更新，全段询问】的线段树。

　　题目要求的是更新指定区间，然后询问全部区间中连续区间的最大值。这题的时限比较紧，最好是用G++来做，C++我的代码超时了。。。。囧！

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cassert>

using namespace std;

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

const int maxn = 16001;
struct Hotel {
    int mx;
    int l, r;

    Hotel(int _set = 0): mx(_set), l(_set), r(_set) {}
};
Hotel seg[maxn << 2];
int late[maxn << 2];

void scan(int &a){
    char ch;

    while ((ch = getchar()) > '9' || ch < '0');
    a = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9'){
        a = a * 10 + ch - '0';
    }
}

void down(int rt, int len) {
    if (~late[rt]) {
        int l = rt << 1;
        int r = rt << 1 | 1;
        int m = len >> 1;

        late[l] = late[r] = late[rt];
        seg[l] = Hotel(late[rt] * (len - m));
        seg[r] = Hotel(late[rt] * m);
        late[rt] = -1;
    }
}

Hotel up(Hotel left, Hotel right, int lLen, int rLen) {
    Hotel ret;

    ret.mx = max(left.mx, right.mx);
    ret.mx = max(ret.mx, left.r + right.l);
    ret.l = left.l;
    ret.r = right.r;
    if (left.l == lLen) ret.l += right.l;
    if (right.r == rLen) ret.r += left.r;

    return ret;
}

void build(int l, int r, int rt) {
    late[rt] = -1;
    if (l == r) {
        seg[rt] = Hotel(1);

        return ;
    }
    int m = (l + r) >> 1;

    build(lson);
    build(rson);
    seg[rt] = up(seg[rt << 1], seg[rt << 1 | 1], m + 1 - l, r - m);
//    printf("@@@%d %d : %d %d %d\n", l, r, seg[rt].l, seg[rt].r, seg[rt].mx);
}

void update(int L, int R, int op, int l, int r, int rt) {
    if (L <= l && r <= R) {
        seg[rt] = Hotel(op * (r - l + 1));
        late[rt] = op;
//        printf("!!!%d %d : %d %d %d\n", l, r, seg[rt].l, seg[rt].r, seg[rt].mx);

        return ;
    }
    int m = (l + r) >> 1;

    down(rt, r - l + 1);
    if (L <= m) update(L, R, op, lson);
    if (m < R) update(L, R, op, rson);
    seg[rt] = up(seg[rt << 1], seg[rt << 1 | 1], m + 1 - l, r - m);
//    printf("!!!%d %d : %d %d %d\n", l, r, seg[rt].l, seg[rt].r, seg[rt].mx);
}

void deal(int len, int n) {
    int op, x, y;

    build(1, len, 1);
    while (n--) {
        scan(op);
        switch (op) {
        case 1: {
            //scanf("%d%d", &x, &y);
            scan(x);
            scan(y);
            update(x, x + y - 1, 0, 1, len, 1);
        }
        break;
        case 2: {
            //scanf("%d%d", &x, &y);
            scan(x);
            scan(y);
            update(x, x + y - 1, 1, 1, len, 1);
        }
        break;
        case 3: {
            printf("%d\n", seg[1].mx);
        }
        break;
        default:
            return ;
        }
    }
}

int main() {
    int n, p;

    while (~scanf("%d%d", &n, &p)) {
        deal(n, p);
    }

    return 0;
}

——written by Lyon