HDU 3308 LCIS

区间合并。给出一些数,记为a[i],两种操作。U x y表示把a[x] 的值改为 y。Q x y表示求xy间的最长连续上升序列(LCIS)。这里的x y都是从0开始的。

其实吧,就是比较当前区间的左右子区间能否相连,也就是说要看a[k] 与 a[k + 1]的大小关系(k表示区间中点)。其他操作跟“正常”题一样。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("
")
#define PI(n) printf("%d
", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s
", s)
#define PSS(s) printf("%s ", n)
#define PC(n) printf("Case %d: ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 9901;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int N = 111111;
const int M = 22;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

int a[N], msum[N << 2], lsum[N << 2], rsum[N << 2], num, n;

void pushup(int rt, int k, int m)
{
    lsum[rt] = lsum[rt << 1];
    rsum[rt] = rsum[rt << 1 | 1];
    msum[rt] = max(msum[rt << 1], msum[rt << 1 | 1]);
    if(a[k] < a[k + 1])
    {
        if(lsum[rt] == (m - (m >> 1)))lsum[rt] += lsum[rt << 1 | 1];
        if(rsum[rt] == (m >> 1))rsum[rt] += rsum[rt << 1];
        msum[rt] = max(msum[rt], lsum[rt << 1 | 1] + rsum[rt << 1]);
    }
}

void build(int l, int r, int rt)
{
    if(l == r)
    {
        RI(a[num++]);
        msum[rt] = rsum[rt] = lsum[rt] = 1;
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    pushup(rt, m, r - l + 1);
}

void update(int p, int x, int l, int r, int rt)
{
    if(l == r)
    {
        a[p] = x;
        return;
    }
    int m = (l + r) >> 1;
    if(p <= m)update(p, x, lson);
    else update(p, x, rson);
    pushup(rt, m, r - l + 1);
}

int query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R)return msum[rt];
    int m = (l + r) >> 1, ans = 0;
    if(L <= m)ans = max(ans, query(L, R, lson));
    if(R > m)ans = max(ans, query(L, R, rson));
    if(a[m] < a[m + 1] && L <= m && m < R)
    {
        int tl = min(rsum[rt << 1], m - L + 1);
        int tr = min(lsum[rt << 1 | 1], R - m);
        ans = max(ans, tl + tr);
    }
    return ans;
}

int main()
{
    //freopen("input.txt", "r", stdin);

    int t, m, x, y;
    char op[5];
    RI(t);
    while(t--)
    {
        RII(n, m);
        num = 1;
        build(1, n, 1);
        while(m--)
        {
            RS(op);
            RII(x, y);
            if(op[0] == 'Q')PI(query(x + 1, y + 1, 1, n, 1));
            else update(x + 1, y, 1, n, 1);
        }
    }
    return 0;
}


原文地址:https://www.cnblogs.com/james1207/p/3315359.html