这题只要想出怎么建树就简单了吧。
他问的是连续k个值最大的是多少
就可以用1~k之和，2~k+1之和，3~k+2之和.......做为结点。
然后就发现，变成了简单的区间更新和区间查询问题了。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 222222
#define MAXM 22222
#define INF 1000000007
#define lch(x) x<<1
#define rch(x) x<<1|1
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int val[MAXN];
int n, k, m;
int t[MAXN];
int mx[4 * MAXN], cover[4 * MAXN];
void up(int rt)
{mx[rt] = max(mx[lch(rt)], mx[rch(rt)]);
}
void down(int rt)
{if(cover[rt]){cover[lch(rt)] += cover[rt];cover[rch(rt)] += cover[rt];mx[lch(rt)] += cover[rt];mx[rch(rt)] += cover[rt];cover[rt] = 0;}
}
void build(int l, int r, int rt)
{cover[rt] = 0;if(l == r){mx[rt] = t[l];return;}int m = (l + r) >> 1;build(lson);build(rson);up(rt);
}
void update(int L, int R, int l, int r, int rt, int v)
{if(L <= l && R >= r){mx[rt] += v;cover[rt] += v;return;}down(rt);int m = (l + r) >> 1;if(m >= L) update(L, R, lson, v);if(m < R) update(L, R, rson, v);up(rt);
}
int query(int L, int R, int l, int r, int rt)
{if(L <= l && R >= r) return mx[rt];down(rt);int tmp = -INF;int m = (l + r) >> 1;if(m >= L) tmp = max(tmp, query(L, R, lson));if(m < R) tmp = max(tmp, query(L, R, rson));return tmp;
}
void change(int x, int y)
{int st = x - k + 1, ed = x;if(st < 1) st = 1;update(st, ed, 1, n, 1, y - val[x]);val[x] = y;
}
int main()
{int T;scanf("%d", &T);while(T--){scanf("%d%d%d", &n, &m, &k);for(int i = 1; i <= n; i++) scanf("%d", &val[i]);t[1] = 0;for(int i = 1; i <= k; i++) t[1] += val[i];for(int i = 2; i <= n - k + 1; i++) t[i] = t[i - 1] - val[i - 1] + val[i - 1 + k];n = n - k + 1;build(1, n, 1);int op, x, y;while(m--){scanf("%d%d%d", &op, &x, &y);if(op == 0) change(x, y);else if(op == 1){int tmp = val[x];change(x, val[y]);change(y, tmp);}else if(op == 2) printf("%d\n", query(x, y - k + 1, 1, n, 1));}}return 0;
}