标签：gis eof inter long lock ack flow segment flush
树套树是什么啊我不知道/dk
我只知道卡常数w
// by Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize( "inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2,-ffast-math,-fsched-spec,unroll-loops,-falign-jumps,-falign-loops,-falign-labels,-fdevirtualize,-fcaller-saves,-fcrossjumping,-fthread-jumps,-funroll-loops,-freorder-blocks,-fschedule-insns,inline-functions,-ftree-tail-merge,-fschedule-insns2,-fstrict-aliasing,-fstrict-overflow,-falign-functions,-fcse-follow-jumps,-fsched-interblock,-fpartial-inlining,no-stack-protector,-freorder-functions,-findirect-inlining,-fhoist-adjacent-loads,-frerun-cse-after-loop,inline-small-functions,-finline-small-functions,-ftree-switch-conversion,-foptimize-sibling-calls,-fexpensive-optimizations,inline-functions-called-once,-fdelete-null-pointer-checks")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using ll = long long;
const int _ = 1 << 21;
struct I {
char fin[_], *p1 = fin, *p2 = fin;
inline char gc() {
return (p1 == p2) && (p2 = (p1 = fin) + fread(fin, 1, _, stdin), p1 == p2) ? EOF : *p1++;
}
inline I& operator>>(int& x) {
bool sign = 1;
char c = 0;
while (c < 48) ((c = gc()) == 45) && (sign = 0);
x = (c & 15);
while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
x = sign ? x : -x;
return *this;
}
inline I& operator>>(double& x) {
bool sign = 1;
char c = 0;
while (c < 48) ((c = gc()) == 45) && (sign = 0);
x = (c - 48);
while ((c = gc()) > 47) x = x * 10 + (c - 48);
if (c == '.') {
double d = 1.0;
while ((c = gc()) > 47) d = d * 0.1, x = x + (d * (c - 48));
}
x = sign ? x : -x;
return *this;
}
inline I& operator>>(char& x) {
do
x = gc();
while (isspace(x));
return *this;
}
inline I& operator>>(string& s) {
s = "";
char c = gc();
while (isspace(c)) c = gc();
while (!isspace(c) && c != EOF) s += c, c = gc();
return *this;
}
} in;
struct O {
char st[100], fout[_];
signed stk = 0, top = 0;
inline void flush() { fwrite(fout, 1, top, stdout), fflush(stdout), top = 0; }
inline O& operator<<(int x) {
if (top > (1 << 20))
flush();
if (x < 0)
fout[top++] = 45, x = -x;
do
st[++stk] = x % 10 ^ 48, x /= 10;
while (x);
while (stk) fout[top++] = st[stk--];
return *this;
}
inline O& operator<<(ll x) {
if (top > (1 << 20))
flush();
if (x < 0)
fout[top++] = 45, x = -x;
do
st[++stk] = x % 10 ^ 48, x /= 10;
while (x);
while (stk) fout[top++] = st[stk--];
return *this;
}
inline O& operator<<(char x) {
fout[top++] = x;
return *this;
}
inline O& operator<<(string s) {
if (top > (1 << 20))
flush();
for (char x : s) fout[top++] = x;
return *this;
}
} out;
#define pb emplace_back
#define fir first
#define sec second
template <class T>
inline void cmax(T& x, const T& y) {
(x < y) && (x = y);
}
template <class T>
inline void cmin(T& x, const T& y) {
(x > y) && (x = y);
}
const int N = 5e4 + 10;
const int mod = 1e9 + 7;
int n, m, a[N], v[N], rt[N], ls[N << 8], rs[N << 8], Cnt = 0;
struct SegMent {
int val[N << 8];
void upd(int& p, int l, int r, int pos, int v) {
if (!p)
p = ++Cnt;
val[p] += v;
if (l == r)
return;
int mid = l + r >> 1;
(pos <= mid) ? upd(ls[p], l, mid, pos, v) : upd(rs[p], mid + 1, r, pos, v);
}
int qry(int p, int a, int b, int l, int r) {
if (!p)
return 0;
if (a <= l && r <= b)
return val[p];
int mid = l + r >> 1, ans = 0;
if (a <= mid)
ans = qry(ls[p], a, b, l, mid);
if (b > mid)
ans += qry(rs[p], a, b, mid + 1, r);
return ans % mod;
}
inline int low(const int& x) { return x & -x; }
inline void upd(int x, int y, int k) {
for (; x <= n; x += low(x)) upd(rt[x], 1, n, y, k);
}
inline ll qry(int l, int r, int a, int b) {
ll ans = 0;
for (; r; r ^= low(r)) {
ans = (ans + qry(rt[r], a, b, 1, n)) % mod;
}
for (--l; l; l ^= low(l)) {
ans = (ans - qry(rt[l], a, b, 1, n) + mod) % mod;
}
return ans;
}
} cnt, qwq;
ll ans = 0;
signed main() {
#ifdef _WIN64
freopen("testdata.in", "r", stdin);
#endif
in >> n >> m;
rep(i, 1, n) in >> a[i] >> v[i];
rep(i, 1, n) cnt.upd(i, a[i], 1), qwq.upd(i, a[i], v[i]);
rep(i, 1, n) {
ans = (ans + cnt.qry(i + 1, n, 1, a[i]) * v[i]) % mod;
ans = (ans + qwq.qry(i + 1, n, 1, a[i])) % mod;
}
while (m--) {
int x, y;
in >> x >> y;
if (x > y)
swap(x, y);
if (x == y) {
out << ans << '\n';
continue;
}
ans = (ans +
(a[x] < a[y] ? 1 : -1) *
(2 * qwq.qry(x, y, min(a[x], a[y]) + 1, max(a[x], a[y]) - 1) +
(cnt.qry(x, y, min(a[x], a[y]) + 1, max(a[x], a[y]) - 1) + 1) * (v[x] + v[y])) %
mod +
mod) %
mod;
ans = (ans + cnt.qry(x, y, 1, min(a[x], a[y]) - 1) * (v[y] - v[x]) % mod + mod) % mod;
ans = (ans + cnt.qry(x, y, max(a[x], a[y]) + 1, n) * (v[x] - v[y]) % mod + mod) % mod;
cnt.upd(x, a[x], -1), qwq.upd(x, a[x], -v[x]);
cnt.upd(y, a[y], -1), qwq.upd(y, a[y], -v[y]);
cnt.upd(x, a[y], 1), qwq.upd(x, a[y], v[y]);
cnt.upd(y, a[x], 1), qwq.upd(y, a[x], v[x]);
swap(a[x], a[y]), swap(v[x], v[y]);
out << ans << '\n';
}
return out.flush(), 0;
}P3759 [TJOI2017]不勤劳的图书管理员 [树套树]
标签：gis eof inter long lock ack flow segment flush
原文地址：https://www.cnblogs.com/Isaunoya/p/12232994.html
