区间合并
按左端点排序,模拟
#include<bits/stdc++.h>
using namespace std;
const int N=1e6+10;
struct node
{
int l,r;
}record[N];
int cmp(node a,node b)
{
return a.l<b.l;
}
int n,l,r;
int main()
{ ios::sync_with_stdio(false);
cin>>n;
int ans=1;
for(int i=0;i<n;i++)
cin>>record[i].l>>record[i].r;
sort(record,record+n,cmp);
int end=record[0].r;
for(int i=1;i<n;i++)
{
if(record[i].l>end)
{
ans++;
end=record[i].r;
}
else if(record[i].r>end)
end=record[i].r;
}
cout<<ans<<endl;
return 0;
}
区间覆盖
左端点排序
贪心方法:每次选当前区间左端点能够与上一个区间连接上的最靠后的右端点,肯定越长越好
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int s, t;
int n;
#define l first
#define r second
typedef pair<int, int> pii;
pii a[N];
int ans;
int main()
{
cin >> s >> t;
cin >> n;
bool ok = false;
for (int i = 0; i < n; i++)
cin >> a[i].l >> a[i].r;
sort(a, a + n);
for (int i = 0; i < n; i++)
{
int j = i;
int end = -0x3f3f3f3f;
while (j < n && a[j].l <= s)
{
end = max(end, a[j].r);
j++;
}
ans++;
if (end >= t)
{
ok = true;
break;
}
if (end < s)
break;
s = end;
i = j - 1;
}
if (ok)
cout << ans << endl;
else
cout << -1 << endl;
return 0;
}
区间选点
右端点排序,因为每次要选每个区间的右端点,因为越靠后相比较而言
越靠后的点会有更大的机会同时存在于后面的区间
#include<bits/stdc++.h>
using namespace std;
int n;
const int N=2e5+10;
struct node
{
int l,r;
}record[N];
int cmp(node a,node b)
{
return a.r<b.r;
}
int main()
{ ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
cin>>n;
for(int i=0;i<n;i++)
{
cin>>record[i].l>>record[i].r;
}
sort(record,record+n,cmp);
int ans=1;
int end=record[0].r;
for(int i=1;i<n;i++)
{
if(record[i].l>end)
{
ans++;
end=record[i].r;
}
}
cout<<ans<<endl;
return 0;
}
最大不相交区间的数量
#include<bits/stdc++.h>
using namespace std;
int n;
const int N=2e5+10;
struct node
{
int l,r;
}record[N];
int cmp(node a,node b)
{
return a.r<b.r;
}
int main()
{ ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
cin>>n;
for(int i=0;i<n;i++)
{
cin>>record[i].l>>record[i].r;
}
sort(record,record+n,cmp);
int ans=1;
int end=record[0].r;
for(int i=1;i<n;i++)
{
if(record[i].l>end)
{
ans++;
end=record[i].r;
}
}
cout<<ans<<endl;
return 0;
}
区间分组
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n;
int a[N], k;
int main()
{
cin >> n;
while (n--)
{
int l, r;
cin >> l >> r;
a[k++] = l * 2;
a[k++] = r * 2 + 1;
}
sort(a, a + k);
int ans = 0;
int maxn = 0;
for (int i = 0; i < k; i++)
{
if (a[i] % 2 == 0)
{
ans++;
}
else
ans--;
maxn = max(maxn, ans);
}
cout << maxn << endl;
return 0;
}
y总做法
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
#define l first
#define r second
typedef pair<int, int> pii;
priority_queue<int, vector<int>, greater<int>> q;
int n;
pii s[N];
int main()
{
cin >> n;
for (int i = 0; i < n; i++)
cin >> s[i].l >> s[i].r;
sort(s, s + n);
for (int i = 0; i < n; i++)
{
pii now = s[i];
if (q.empty() || q.top() >= now.l)
q.push(now.r);
else
{
q.pop();
q.push(now.r);
}
}
cout << q.size() << endl;
return 0;
}