UESTC 1425 Another LCIS

UESTC_1425

    更多和区间合并相关的线段树问题可以参考胡浩的博客：http://www.notonlysuccess.com/index.php/segment-tree-complete/。

    这个题和HDU_3308很像，只不过要多处理一下区间加和的操作，为了能够方便的修改、获取左区间的右端点和右区间左端点的大小关系，可以用两个标记lx[]、rx[]分别表示当前区间左端点以及右端点的值。

#include<stdio.h>
#include<string.h>
#define MAXD 100010
int N, Q, a[MAXD], lc[4 * MAXD], rc[4 * MAXD], lx[4 * MAXD], rx[4 * MAXD], mc[4 * MAXD], add[4 * MAXD];
int getmax(int x, int y)
{
    return x > y ? x : y;
}
void update(int cur, int x, int y)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    mc[cur] = getmax(mc[ls], mc[rs]);
    lc[cur] = lc[ls], rc[cur] = rc[rs];
    if(rx[ls] < lx[rs])
    {
        if(rc[ls] + lc[rs] > mc[cur])
            mc[cur] = rc[ls] + lc[rs];
        if(lc[ls] == mid - x + 1)
            lc[cur] += lc[rs];
        if(rc[rs] == y - mid)
            rc[cur] += rc[ls];
    }
    lx[cur] = lx[ls], rx[cur] = rx[rs];
}
void pushdown(int cur)
{
    int ls = cur << 1, rs = (cur << 1) | 1;
    if(add[cur])
    {
        add[ls] += add[cur], add[rs] += add[cur];
        lx[ls] += add[cur], rx[ls] += add[cur];
        lx[rs] += add[cur], rx[rs] += add[cur];
        add[cur] = 0;
    }
}
void build(int cur, int x, int y)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    add[cur] = 0;
    if(x == y)
    {
        mc[cur] = lc[cur] = rc[cur] = 1;
        lx[cur] = rx[cur] = a[x];
        return ;
    }
    build(ls, x, mid);
    build(rs, mid + 1, y);
    update(cur, x, y);
}
void init()
{
    int i, j, k;
    scanf("%d%d", &N, &Q);
    for(i = 1; i <= N; i ++)
        scanf("%d", &a[i]);
    build(1, 1, N);
}
int query(int cur, int x, int y, int s, int t, int fa, int &ans)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    if(x >= s && y <= t)
    {
        if(mc[cur] > ans)
            ans = mc[cur];
        return fa == -1 ? lc[cur] : rc[cur];
    }
    pushdown(cur);
    if(mid >= t)
        return query(ls, x, mid, s, t, -1, ans);
    else if(mid + 1 <= s)
        return query(rs, mid + 1, y, s, t, 1, ans);
    else
    {
        int ln = query(ls, x, mid, s, t, 1, ans), rn = query(rs, mid + 1, y, s, t, -1, ans);
        if(rx[ls] < lx[rs])
        {
            if(ln + rn > ans)
                ans = ln + rn;
            if(fa == -1)
                return lc[ls] == mid - x + 1 ? lc[ls] + rn : lc[ls];
            else
                return rc[rs] == y - mid ? rc[rs] + ln : rc[rs];
        }
        return fa == -1 ? lc[ls] : rc[rs];
    }
}
void refresh(int cur, int x, int y, int s, int t, int v)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    if(x >= s && y <= t)
    {
        add[cur] += v, lx[cur] += v, rx[cur] += v;
        return ;
    }
    pushdown(cur);
    if(mid >= s)
        refresh(ls, x, mid, s, t, v);
    if(mid + 1 <= t)
        refresh(rs, mid + 1, y, s, t, v);
    update(cur, x, y);
}
void solve()
{
    int i, j, k, x, y, z, ans;
    char b[5];
    for(i = 0; i < Q; i ++)
    {
        scanf("%s", b);
        if(b[0] == 'q')
        {
            scanf("%d%d", &x, &y);
            ans = 0;
            query(1, 1, N, x, y, -1, ans);
            printf("%d\n", ans);
        }
        else
        {
            scanf("%d%d%d", &x, &y, &z);
            refresh(1, 1, N, x, y, z);
        }
    }
}
int main()
{
    int t, tt;
    scanf("%d", &t);
    for(tt = 0; tt < t; tt ++)
    {
        init();
        printf("Case #%d:\n", tt + 1);
        solve();
    }
    return 0;
}