1 条题解
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1
考虑二分,若 为当前二分到的数,则把 的数设成 ,另外的设成 ,升序排序相当于把 移到最前面,降序相当于把 移动到最后面(再该区间范围内)。
还是很简单的 QwQ。
code:
#include<bits/stdc++.h> using namespace std; inline int read() { register char in = getchar(),f = 0;register int t = 0; for(;!isdigit(in);in = getchar())if(!(in ^ 45)) f = 1; for(;in >= '0' && in <= '9';in = getchar()) t = (t << 1) + (t << 3) + (in ^ 48); return f ? -t : t; } inline void write(register int x) { static int t[25];register int tp = 0; if(x == 0) return (void) (puts("0")); else if(x < 0) putchar('-'),x = -x; while(x) t[++tp] = x % 10,x /= 10; while(tp) putchar(t[tp--] + '0'); puts(""); } #define stdi stdin #define stdo stdout int n = read(),m = read(),a[100012],ans,q; struct input { int k,l,r; }s[100012]; struct node { int l,r; mutable int v; node(int l,int r = 0,int v = 0) : l(l),r(r),v(v){} bool operator < (const node &other)const { return l < other.l; } };set<node> tree; inline set<node> :: iterator split(register int pos) { register set<node> :: iterator it = tree.lower_bound(node(pos)); if(it != tree.end() && it -> l == pos) return it; it--; if(it -> r < pos) return tree.end(); register int l = it -> l; register int r = it -> r; register int v = it -> v; tree.erase(it); tree.insert(node(l,pos - 1,v)); return tree.insert(node(pos,r,v)).first; } inline int find(register int l,register int r) { register int ret = 0; set<node> :: iterator rr = split(r + 1),ll = split(l),it; for(it = ll;it != rr;it++) { if(it -> v) ret += it -> r - it -> l + 1; } return ret; } void amend(register int l,register int r,register int k) { set<node> :: iterator rr = split(r + 1),ll = split(l); tree.erase(ll,rr),tree.insert(node(l,r,k)); return; } inline short check(int mid) { tree.clear(); for(int i = 1;i <= n;i++) tree.insert(node(i,i,a[i] >= mid)); for(int i = 1;i <= m;i++) { register int op = s[i].k,l = s[i].l,r = s[i].r,M = find(l,r); if(op == 0) amend(l,r - M,0),amend(r - M + 1,r,1); if(op == 1) amend(l,l + M - 1,1),amend(l + M,r,0); } return split(q) -> v == 1; } int main() { setvbuf(stdi,(char*)calloc(1 << 20,sizeof(char)),_IOFBF,1 << 20); setvbuf(stdo,(char*)calloc(1 << 20,sizeof(char)),_IOFBF,1 << 20); for(int i = 1;i <= n;i++) a[i] = read(); for(int i = 1;i <= m;i++) { s[i].k = read(); s[i].l = read(); s[i].r = read(); } q = read(); int l = 1,r = n,mid; while(l <= r) { mid = l + r >> 1; if(check(mid)) ans = mid,l = mid + 1; else r = mid - 1; } write(ans); }
信息
- ID
- 15338
- 时间
- 6000ms
- 内存
- 256MiB
- 难度
- 10
- 标签
- 递交数
- 1
- 已通过
- 1
- 上传者