題意
Sol
ODT板子題。
操作1直接拆區間就行。
#include<bits/stdc++.h>
#define fi first
#define se second
const int MAXN = 2e5 + 10;
using namespace std;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < `0` || c > `9`) {if(c == `-`) f = -1; c = getchar();}
while(c >= `0` && c <= `9`) x = x * 10 + c - `0`, c = getchar();
return x * f;
}
int N, M;
#define sit set<Node>::iterator
struct Node {
int l, r;
mutable int v;
bool operator < (const Node &rhs) const {
return l < rhs.l;
}
};
set<Node> s;
sit split(int p) {
sit pos = s.lower_bound({p, 0, 0});
if(pos->l == p) return pos;
pos--;
int L = pos->l, R = pos->r, V = pos->v;
s.erase(pos);
s.insert({L, p - 1, V});
return s.insert({p, R, V}).fi;
}
void Mem(int l, int r) {
sit ed = split(r + 1), bg = split(l);
s.erase(bg, ed);
s.insert({l, r, 0});
}
void Fix(int l0, int r0, int l, int r) {
int num = 0;
sit ed = split(r + 1), bg = split(l);
for(sit i = bg; i != ed; i++) if(i->v == 1) num += i->r - i->l + 1;
s.erase(bg, ed);
s.insert({l, r, 0});
ed = split(r0 + 1), bg = split(l0);
sit gg; int RR = -1;
for(sit i = bg; i != ed; i++) {
if(i -> v == 1) continue;
if(num <= 0) return ;
int len = i->r - i->l + 1;
if(len <= num) {i -> v = 1, num -= len; gg = i; RR = i->r; continue;}
int L = i->l, R = i->r;
s.erase(i);
s.insert({L, L + num - 1, 1});
s.insert({L + num, R, 0});
return ;
}
if(RR > l0) {//Ò»¸ö²¢Ã»ÓÐʲôÂÑÓõÄÓÅ»¯
gg++;
s.erase(bg, gg);
s.insert({l0, RR, 1});
}
}
int Query(int l, int r) {
sit ed = split(r + 1), bg = split(l);
int pre = 0, ans = 0;
for(sit i = bg; i != ed; i++) {
if(i->v == 0) ans = max(ans, i->r - i->l + 1 + pre), pre += i->r - i->l + 1;
else pre = 0;
}
return ans;
}
int main() {
//freopen("a.in", "r", stdin);
N = read(); M = read();
for(int i = 1; i <= N; i++) s.insert({i, i, 1});
s.insert({N + 1, N + 1, 1});
for(int i = 1; i <= M; i++) {
int opt = read(), l = read(), r = read();
if(opt == 0) Mem(l, r);
else if(opt == 1) {
int l1 = read(), r1 = read();
Fix(l1, r1, l, r);
} else printf("%d
", Query(l, r));
}
return 0;
}