题目大意:一个全排列,两种操作:
1. $0\;l\;r:$把$[l,r]$升序排序
2. $1\;l\;r:$把$[l,r]$降序排序
最后询问第$k$位是什么
题解:二分答案,把比这个数大的赋成$1$,否则为$0$,线段树区间和和区间赋$01$,最后判断第$k$位是$0$是$1$,若为$1$则还可以变大,否则变小
卡点:修改后没有$update$
C++ Code:
#include <cstdio>
#include <cctype>
namespace __R {
int x, ch;
inline int read() {
ch = getchar();
while (isspace(ch)) ch = getchar();
for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
return x;
}
}
using __R::read;
#define maxn 100010
int n, m, q;
int s[maxn], op[maxn], L[maxn], R[maxn];
namespace SgT {
int V[maxn << 2], tg[maxn << 2];
int num, L, R;
void build(int rt, int l, int r) {
tg[rt] = -1;
if (l == r) {
V[rt] = s[l] > num;
return ;
}
int mid = l + r >> 1;
build(rt << 1, l, mid);
build(rt << 1 | 1, mid + 1, r);
V[rt] = V[rt << 1] + V[rt << 1 | 1];
}
void build(int __num) {
num = __num;
build(1, 1, n);
}
inline void pushdown(int rt, int len) {
int &__tg = tg[rt];
V[rt << 1] = (len + 1 >> 1) * __tg;
tg[rt << 1] = __tg;
V[rt << 1 | 1] = (len >> 1) * __tg;
tg[rt << 1 | 1] = __tg;
__tg = -1;
}
int __query(int rt, int l, int r) {
if (L <= l && R >= r) return V[rt];
int mid = l + r >> 1, ans = 0;
if (~tg[rt]) pushdown(rt, r - l + 1);
if (L <= mid) ans = __query(rt << 1, l, mid);
if (R > mid) ans += __query(rt << 1 | 1, mid + 1, r);
return ans;
}
int query(int __L, int __R) {
L = __L, R = __R;
return __query(1, 1, n);
}
void __modify(int rt, int l, int r) {
if (L <= l && R >= r) {
V[rt] = num * (r - l + 1);
tg[rt] = num;
return ;
}
int mid = l + r >> 1;
if (~tg[rt]) pushdown(rt, r - l + 1);
if (L <= mid) __modify(rt << 1, l, mid);
if (R > mid) __modify(rt << 1 | 1, mid + 1, r);
V[rt] = V[rt << 1] + V[rt << 1 | 1];
}
void modify(int __L, int __R, int __num) {
L = __L, R = __R, num = __num;
__modify(1, 1, n);
}
}
bool check(int mid) {
SgT::build(mid);
for (int i = 1; i <= m; i++) {
int _1 = SgT::query(L[i], R[i]), _0 = R[i] - L[i] + 1 - _1;
if (op[i]) {
SgT::modify(L[i], L[i] + _1 - 1, 1);
SgT::modify(L[i] + _1, R[i], 0);
} else {
SgT::modify(L[i], L[i] + _0 - 1, 0);
SgT::modify(L[i] + _0, R[i], 1);
}
}
return SgT::query(q, q);
}
int main() {
n = read(), m = read();
for (int i = 1; i <= n; i++) s[i] = read();
for (int i = 1; i <= m; i++) {
op[i] = read(), L[i] = read(), R[i] = read();
}
q = read();
int l = 1, r = n, ans = 1;
while (l <= r) {
int mid = l + r >> 1;
if (check(mid)) l = mid + 1;
else {
r = mid - 1;
ans = mid;
}
}
printf("%d\n", ans);
return 0;
}
[洛谷P2824][HEOI2016/TJOI2016]排序
原文:https://www.cnblogs.com/Memory-of-winter/p/9978008.html