這題只要想出怎麼建樹就簡單了吧。
他問的是連續k個值最大的是多少
就可以用1~k之和，2~k+1之和，3~k+2之和.......做為結點。
然後就發現，變成了簡單的區間更新和區間查詢問題了。
[cpp]
#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;
}
#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;
}
