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2 years ago
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
vector<PII> segs;
vector<PII> res;
void merge() {
if (!segs.size()) return; // 本题中没有实际意义为了防止下面segs[0]越界
// ① 按左端点排序
sort(segs.begin(), segs.end());
int st = segs[0].first, ed = segs[0].second; // 第一只猴子默认为大王
for (int i = 1; i < segs.size(); i++) // 后续的猴子开始枚举
if (ed < segs[i].first) { // ② 无交集
res.push_back({st, ed}); // 发现新大陆
st = segs[i].first, ed = segs[i].second; // 重新设置起始、终止值
} else // ③ 有交集
ed = max(ed, segs[i].second); // PK最大值
// ③ 残余势力入数组
res.push_back({st, ed});
}
int main() {
int n;
cin >> n;
while (n--) {
int l, r;
cin >> l >> r;
segs.push_back({l, r});
}
merge();
printf("%d\n", res.size());
return 0;
}