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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判断一下是否满足递增即可。 #include
#include
#include
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; }
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