题目链接:hdu 3308 LCIS
题目大意:给定一个序列,两种操作:
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 <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;
}