hdu 3308 LCIS(线段树)

Posted 2020-06-13

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pid=3308" target="_blank" style="">题目链接：hdu 3308 LCIS

题目大意：给定一个序列，两种操作：

• Q l r：查询区间l，r中的最长连续递增序列长度
• U p x：将位置p上的数改成x

解题思路：线段树上的区间合并，这是在左右子树合并的时候要推断一下是否满足递增就可以。

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

using namespace std;

const int maxn = 1e5 + 5;

int N, M, a[maxn];

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2];
int L[maxn << 2], R[maxn << 2], S[maxn << 2];

void pushup (int u) {
    int mid = (lc[u] + rc[u]) / 2;
    S[u] = max(max(S[lson(u)], S[rson(u)]), (a[mid] < a[mid+1] ? R[lson(u)] + L[rson(u)] : 0));
    L[u] = L[lson(u)] + (L[lson(u)] == rc[lson(u)] - lc[lson(u)] + 1 && a[mid] < a[mid + 1] ?
 L[rson(u)] : 0);
    R[u] = R[rson(u)] + (R[rson(u)] == rc[rson(u)] - lc[rson(u)] + 1 && a[mid] < a[mid + 1] ?
 R[lson(u)] : 0);
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;

    if (l == r) {
        L[u] = R[u] = S[u] = 1;
        return;
    }

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int x, int v) {
    if (lc[u] == x && rc[u] == x) {
        a[x] = v;
        return;
    }

    int mid = (lc[u] + rc[u]) / 2;
    if (x <= mid)
        modify(lson(u), x, v);
    else
        modify(rson(u), x, v);
    pushup(u);
}

int query (int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r)
        return S[u];

    int mid =(lc[u] + rc[u]) / 2, ret;
    if (r <= mid)
        ret = query(lson(u), l, r);
    else if (l > mid)
        ret = query(rson(u), l, r);
    else {
        int ll = query(lson(u), l, r);
        int rr = query(rson(u), l, r);

        int A = min(R[lson(u)], mid - l + 1);
        int B = min(L[rson(u)], r - mid);

        ret = max(max(ll, rr), a[mid] < a[mid + 1] ?
 A + B : 0);
    }
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d%d", &N, &M);
        for (int i = 0; i < N; i++)
            scanf("%d", &a[i]);

        build(1, 0, N-1);

        int l, r;
        char op[5];
        while (M--) {
            scanf("%s%d%d", op, &l, &r);
            if (op[0] == ‘U‘)
                modify(1, l, r);
            else
                printf("%d\n", query(1, l, r));
        }
    }
    return 0;
}