HDU 1540 Tunnel Warfare

线段树求一个点所处区间的最大连续长度，用lsum记录区间的左连续长度，rsum记录区间的

右连续长度，查询时判断p点处于左子树还是右子树，然后计算左子树的右连续长度+右子树

的左连续长度的和。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1

const int MAXN = 50050;
int lsum[MAXN << 2], rsum[MAXN << 2];
int stack[MAXN];
int n, m;

void build(int l, int r, int rt)
{
    int mid = l + r >> 1;
    lsum[rt] = rsum[rt] = r - l + 1;
    if(l == r) return;
    build(lson);
    build(rson);
}

void PushUp(int l, int r, int rt)
{
    int mid = l + r >> 1;
    lsum[rt] = lsum[rt << 1];
    rsum[rt] = rsum[rt << 1 | 1];
    if(lsum[rt << 1] == mid - l + 1)
        lsum[rt] += lsum[rt << 1 | 1];
    if(rsum[rt << 1 | 1] == r - mid)
        rsum[rt] += rsum[rt << 1];
}

void update(int c, int p, int l, int r, int rt)
{
    int mid = l + r >> 1;
    if(p <= l && r <= p)
    {
        lsum[rt] = rsum[rt] = c;
        return;
    }
    if(p <= mid)
        update(c, p, lson);
    else
        update(c, p, rson);
    PushUp(l, r, rt);
}

int query(int p, int l, int r, int rt)
{
    int mid = l + r >> 1;
    if(p <= l && r <= p)
    {
        return lsum[rt];
    }
    if(p <= mid)
    {
        if(mid - rsum[rt << 1] + 1 <= p) //判断p所处的位置
        {
            return rsum[rt << 1] + lsum[rt << 1 | 1];
        }
        else
            return query(p, lson);
    }
    else
    {
        if(mid + 1 + lsum[rt << 1 | 1] - 1 >= p)
        {
            return rsum[rt << 1] + lsum[rt << 1 | 1];
        }
        else
            return query(p, rson);
    }
}

void operation()
{
    int top = 0, x;
    char op[5];
    while(m --)
    {
        scanf("%s", op);
        if('D' == op[0])
        {
            scanf("%d", &x);
            stack[++ top] = x;
            update(0, x, 1, n, 1);
        }
        else if('Q' == op[0])
        {
            scanf("%d", &x);
            printf("%d\n", query(x, 1, n, 1));
        }
        else
        {
            x = stack[top --];
            update(1, x, 1, n, 1);
        }
    }
}

int main()
{

    while(scanf("%d%d", &n, &m) == 2)
    {
        build(1, n, 1);
        operation();
    }
    return 0;
}