仔细想想这个并不是什么正常的线段树,因为它除了PushDow有一部分信息是从上往下传递的。
http://acm.hdu.edu.cn/showproblem.php?pid=5306
1.区间更新min
2.区间查询max
3.区间查询sum
struct SegmentTreeBeats {
#define ls (u << 1)
#define rs (u << 1 | 1)
#define mid ((l + r) >> 1)
static const int MAXN = 2e5 + 10;
static const int INF = 0x3f3f3f3f;
int tmi[MAXN << 2];
ll sum[MAXN << 2];
int mx1[MAXN << 2], cmx[MAXN << 2], mx2[MAXN << 2];
void PushUp(int u) {
sum[u] = sum[ls] + sum[rs];
if(mx1[ls] < mx1[rs]) {
mx1[u] = mx1[rs], cmx[u] = cmx[rs];
mx2[u] = max(mx1[ls], mx2[rs]);
} else if (mx1[ls] > mx1[rs]) {
mx1[u] = mx1[ls], cmx[u] = cmx[ls];
mx2[u] = max(mx2[ls], mx1[rs]);
} else {
mx1[u] = mx1[ls], cmx[u] = cmx[ls] + cmx[rs];
mx2[u] = max(mx2[ls], mx2[rs]);
}
}
void PushTagMin(int u, int t) {
if(mx1[u] <= t)
return;
sum[u] += 1LL * (t - mx1[u]) * cmx[u];
mx1[u] = tmi[u] = t;
}
void PushDown(int u, int l, int r) {
int t = tmi[u];
if(t != INF) {
PushTagMin(ls, t);
PushTagMin(rs, t);
tmi[u] = INF;
}
}
void Build(int u, int l, int r) {
tmi[u] = INF;
if(l == r) {
sum[u] = mx1[u] = a[l];
cmx[u] = 1, mx2[u] = -INF;
return;
}
Build(ls, l, mid);
Build(rs, mid + 1, r);
PushUp(u);
}
void UpdateMin(int u, int l, int r, int L, int R, int v) {
if(mx1[u] <= v)
return;
if(L <= l && r <= R && mx2[u] < v) {
PushTagMin(u, v);
return;
}
PushDown(u, l, r);
if(L <= mid) UpdateMin(ls, l, mid, L, R, v);
if(R >= mid + 1) UpdateMin(rs, mid + 1, r, L, R, v);
PushUp(u);
}
ll QuerySum(int u, int l, int r, int L, int R) {
if (L <= l && r <= R)
return sum[u];
PushDown(u, l, r);
ll res = 0;
if(L <= mid) res += QuerySum(ls, l, mid, L, R);
if(R >= mid + 1) res += QuerySum(rs, mid + 1, r, L, R);
return res;
}
int QueryMax(int u, int l, int r, int L, int R) {
if(L <= l && r <= R)
return mx1[u];
PushDown(u, l, r);
int res = -INF;
if(L <= mid) res = max(res, QueryMax(ls, l, mid, L, R));
if(R >= mid + 1) res = max(res, QueryMax(rs, mid + 1, r, L, R));
return res;
}
#undef ls
#undef rs
#undef mid
} stb;
用势能分析法可以得到m次操作总共是 (O(mlog n)) ?吉老师似乎说这个有问题,但是构造不出卡掉他的数据。
混合标记:
1.区间加法
2.区间取min
3.区间取max
4.区间查询sum
5.区间查询min
6.区间查询max
没有在OJ上面验证过哦,只能和另一个暴力程序对拍了。
struct SegmentTreeBeats {
#define ls (u << 1)
#define rs (u << 1 | 1)
#define mid ((l + r) >> 1)
static const int MAXN = 2e5 + 10;
static const int INF = 0x3f3f3f3f;
int tad[MAXN << 2], tmi[MAXN << 2], tmx[MAXN << 2];
ll sum[MAXN << 2];
int mi1[MAXN << 2], cmi[MAXN << 2], mi2[MAXN << 2];
int mx1[MAXN << 2], cmx[MAXN << 2], mx2[MAXN << 2];
void PushUp(int u) {
sum[u] = sum[ls] + sum[rs];
if(mi1[ls] < mi1[rs]) {
mi1[u] = mi1[ls], cmi[u] = cmi[ls];
mi2[u] = min(mi2[ls], mi1[rs]);
} else if(mi1[ls] > mi1[rs]) {
mi1[u] = mi1[rs], cmi[u] = cmi[rs];
mi2[u] = min(mi1[ls], mi2[rs]);
} else {
mi1[u] = mi1[ls], cmi[u] = cmi[ls] + cmi[rs];
mi2[u] = min(mi2[ls], mi2[rs]);
}
if(mx1[ls] < mx1[rs]) {
mx1[u] = mx1[rs], cmx[u] = cmx[rs];
mx2[u] = max(mx1[ls], mx2[rs]);
} else if (mx1[ls] > mx1[rs]) {
mx1[u] = mx1[ls], cmx[u] = cmx[ls];
mx2[u] = max(mx2[ls], mx1[rs]);
} else {
mx1[u] = mx1[ls], cmx[u] = cmx[ls] + cmx[rs];
mx2[u] = max(mx2[ls], mx2[rs]);
}
}
void PushTagAdd(int u, int l, int r, int t) {
sum[u] += 1LL * (r - l + 1) * t;
tad[u] += t, mi1[u] += t, mx1[u] += t;
if(mi2[u] != INF) mi2[u] += t;
if(tmi[u] != INF) tmi[u] += t;
if(mx2[u] != -INF) mx2[u] += t;
if(tmx[u] != -INF) tmx[u] += t;
}
void PushTagMin(int u, int t) {
if(mx1[u] <= t)
return;
sum[u] += 1LL * (t - mx1[u]) * cmx[u];
if(mi1[u] == mx1[u]) mi1[u] = t;
if(mi2[u] == mx1[u]) mi2[u] = t;
if(tmx[u] > t) tmx[u] = t;
mx1[u] = tmi[u] = t;
}
void PushTagMax(int u, int t) {
if(mi1[u] >= t)
return;
sum[u] += 1LL * (t - mi1[u]) * cmi[u];
if(mx1[u] == mi1[u]) mx1[u] = t;
if(mx2[u] == mi1[u]) mx2[u] = t;
if(tmi[u] < t) tmi[u] = t;
mi1[u] = tmx[u] = t;
}
void PushDown(int u, int l, int r) {
int t = tad[u];
if(t != 0) {
PushTagAdd(ls, l, mid, t);
PushTagAdd(rs, mid + 1, r, t);
tad[u] = 0;
}
t = tmi[u];
if(t != INF) {
PushTagMin(ls, t);
PushTagMin(rs, t);
tmi[u] = INF;
}
t = tmx[u];
if(t != -INF) {
PushTagMax(ls, t);
PushTagMax(rs, t);
tmx[u] = -INF;
}
}
void Build(int u, int l, int r) {
tad[u] = 0, tmi[u] = INF, tmx[u] = -INF;
if(l == r) {
sum[u] = mi1[u] = mx1[u] = a[l];
cmi[u] = 1, mi2[u] = INF;
cmx[u] = 1, mx2[u] = -INF;
return;
}
Build(ls, l, mid);
Build(rs, mid + 1, r);
PushUp(u);
}
void UpdateAdd(int u, int l, int r, int L, int R, int v) {
if(L <= l && r <= R) {
PushTagAdd(u, l, r, v);
return;
}
PushDown(u, l, r);
if(L <= mid) UpdateAdd(ls, l, mid, L, R, v);
if(R >= mid + 1) UpdateAdd(rs, mid + 1, r, L, R, v);
PushUp(u);
}
void UpdateMin(int u, int l, int r, int L, int R, int v) {
if(mx1[u] <= v)
return;
if(L <= l && r <= R && mx2[u] < v) {
PushTagMin(u, v);
return;
}
PushDown(u, l, r);
if(L <= mid) UpdateMin(ls, l, mid, L, R, v);
if(R >= mid + 1) UpdateMin(rs, mid + 1, r, L, R, v);
PushUp(u);
}
void UpdateMax(int u, int l, int r, int L, int R, int v) {
if(mi1[u] >= v)
return;
if(L <= l && r <= R && mi2[u] > v) {
PushTagMax(u, v);
return;
}
PushDown(u, l, r);
if(L <= mid) UpdateMax(ls, l, mid, L, R, v);
if(R >= mid + 1) UpdateMax(rs, mid + 1, r, L, R, v);
PushUp(u);
}
ll QuerySum(int u, int l, int r, int L, int R) {
if (L <= l && r <= R)
return sum[u];
PushDown(u, l, r);
ll res = 0;
if(L <= mid) res += QuerySum(ls, l, mid, L, R);
if(R >= mid + 1) res += QuerySum(rs, mid + 1, r, L, R);
return res;
}
int QueryMin(int u, int l, int r, int L, int R) {
if(L <= l && r <= R)
return mi1[u];
PushDown(u, l, r);
int res = INF;
if(L <= mid) res = min(res, QueryMin(ls, l, mid, L, R));
if(R >= mid + 1) res = min(res, QueryMin(rs, mid + 1, r, L, R));
return res;
}
int QueryMax(int u, int l, int r, int L, int R) {
if(L <= l && r <= R)
return mx1[u];
PushDown(u, l, r);
int res = -INF;
if(L <= mid) res = max(res, QueryMax(ls, l, mid, L, R));
if(R >= mid + 1) res = max(res, QueryMax(rs, mid + 1, r, L, R));
return res;
}
#undef ls
#undef rs
#undef mid
} stb;
吉老师说 (O(mlog^2 n))。