# CS Academy Round #72. Line Enemies

,

https://csacademy.com/contest/round-72/task/enemy/

## solution

• 同じ区間のとき$0$
• ふたつが交わったないとき$1$
• どちらとも交わらない区間があるとき$2$
• $[L_1, R_1]$とだけ交わる区間および$[L_1, R_1]$とだけ交わる区間があるとき$3$
• それ以外は$-1$

## implementation

#include <bits/stdc++.h>
using namespace std;

int main() {
// prepare
set<pair<int, int> > s;
map<int, int> lcnt, rcnt;

// query
int query; scanf("%d", &query);
while (query --) {
int type; scanf("%d", &type);
if (type == 1) {
int l, r; scanf("%d%d", &l, &r); -- l;
s.emplace(l, r);
lcnt[l] += 1;
rcnt[r] += 1;
} else if (type == 2) {
int l, r; scanf("%d%d", &l, &r); -- l;
s.erase(make_pair(l, r));
if (not (lcnt[l] -= 1)) lcnt.erase(l);
if (not (rcnt[r] -= 1)) rcnt.erase(r);
} else if (type == 3) {
int l1, r1; scanf("%d%d", &l1, &r1); -- l1;
int l2, r2; scanf("%d%d", &l2, &r2); -- l2;
bool included = (l1 <= l2 and r2 <= r1) or (l2 <= l1 and r1 <= r2);
int lmax = lcnt.rbegin()->first;
int rmin = rcnt. begin()->first;
int result;
if (l1 == l2 and r1 == r2) {
result = 0;
} else if (r1 <= l2 or r2 <= l1) {
result = 1;
} else if (max(r1, r2) <= lmax or rmin <= min(l1, l2)) {
result = 2;
} else if (not included and rmin <= max(l1, l2) and rmin <= lmax and min(r1, r2) <= lmax) {
result = 3;
} else {
result = -1;
}
printf("%d\n", result);
} else {
assert (false);
}
}
return 0;
}