PKU 4047 - Garden (线段树 + 区间修改)

题意

给出三个操作。单点更新,交换两个点,区间询问。

询问的时候要求输出区间内k个值的最大和。

思路

一开始想着维护k个值的最大和,没思路。

后来想到可以把k个值压缩成一个点,这样就变成普通的线段树了。

这样的话,当修改一个点时,影响的新的区间是l = max(1, n - k + 1), r = min(N - k + 1, n).

交换,单点更新都可以看成某个区间内增加一个值。

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <ctime>
#include <cstdlib>
#include <fstream>
#include <string>
#include <sstream>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const LL MAXN = 2e5 + 10;
const int MOD = 20071027;
 
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
 
int num[MAXN], n, m, k;
vector<int> sum;
 
struct SEGTREE
{
    int add, nmax;
}segt[MAXN << 2];
 
void PushUp(int rt)
{
    segt[rt].nmax = max(segt[LRT].nmax, segt[RRT].nmax);
}
 
void PushDown(int rt)
{
    const int &add = segt[rt].add;
    segt[LRT].nmax += add;
    segt[RRT].nmax += add;
    segt[LRT].add += add;
    segt[RRT].add += add;
    segt[rt].add = 0;
}
 
int Query(int rt, int l, int r, int L, int R)
{
    if (L <= l && r <= R) return segt[rt].nmax;
    if (segt[rt].add) PushDown(rt);
    int mid = MID(l, r);
    int ret = -INF;
    if (L <= mid) ret = max(ret, Query(LC, L, R));
    if (R > mid) ret = max(ret, Query(RC, L, R));
    return ret;
}
 
void Update(int rt, int l, int r, int L, int R, int val)
{
    if (L <= l && r <= R)
    {
        segt[rt].add += val;
        segt[rt].nmax += val;
        return;
    }
    if (segt[rt].add) PushDown(rt);
    int mid = MID(l, r);
    if (L <= mid) Update(LC, L, R, val);
    if (R > mid) Update(RC, L, R, val);
    PushUp(rt);
}
 
void Build(int rt, int l, int r)
{
    segt[rt].add = 0;
    if (l == r)
    {
        segt[rt].nmax = sum[l];
        return;
    }
    int mid = MID(l, r);
    Build(LC); Build(RC);
    PushUp(rt);
}
 
void Solve()
{
    const int N = n - k + 1;
    int a, b, c;
    while (m--)
    {
        scanf("%d%d%d", &a, &b, &c);
        if (a == 0)
        {
            int val = c - num[b];
            num[b] = c;
            Update(1, 1, N, max(1, b - k + 1), min(b, N), val);
        }
        else if (a == 1)
        {
            int val = num[c] - num[b];
            Update(1, 1, N, max(1, b - k + 1), min(b, N), val);
            val = num[b] - num[c];
            swap(num[b], num[c]);
            Update(1, 1, N, max(1, c - k + 1), min(c, N), val);
        }
        else printf("%d\n", Query(1, 1, N, b, c - k + 1));
    }
}
 
int main()
{
    //ROP;
    int T, i, j;
    scanf("%d", &T);
    while (T--)
    {
        sum.clear(); sum.PB(9999); sum.PB(0);
        scanf("%d%d%d", &n, &m, &k);
        int temp = 0;
        for (i = 1; i <= n; i++)
        {
            scanf("%d", #[i]);
            temp += num[i];
            if (i > k)
            {
                temp -= num[i - k];
                sum.PB(temp);
            }
            else sum.back() = temp;
        }
        Build(1, 1, n - k + 1);
        Solve();
    }
    return 0;
}

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