集合帖：区间问题

一、AcWing 803：区间合并
（1）题目来源：https://www.acwing.com/problem/content/805/
（2）算法代码：https://blog.csdn.net/hnjzsyjyj/article/details/145067059

#include <bits/stdc++.h>
using namespace std;const int maxn=1e5+5;
struct Section {int x;int y;
} a[maxn];bool cmp(Section u,Section v) {if(u.x==v.x) return u.y<v.y;return u.x<v.x;
}int main() {int n;cin>>n;for(int i=1; i<=n; i++) {cin>>a[i].x>>a[i].y;}sort(a+1,a+1+n,cmp);int base=a[1].y;int cnt=1;for(int i=1; i<=n; i++) {if(a[i].x<=base) base=max(base,a[i].y);else cnt++,base=a[i].y;}cout<<cnt<<endl;return 0;
}/*
in:
10
-15 -8
-20 23
-2 11
2 22
18 23
11 27
-8 21
-18 14
-17 -12
-23 8
out:
1
*/

二、洛谷 P1496：火烧赤壁
（1）题目来源：https://www.luogu.com.cn/problem/P1496
（2）算法代码：https://blog.csdn.net/hnjzsyjyj/article/details/145073398

#include <bits/stdc++.h>
using namespace std;const int maxn=2e5;
struct Point {int x;int y;
} a[maxn];bool cmp(Point u,Point v) {if(u.x==v.x) return u.y<v.y;return u.x<v.x;
}int main() {int n;cin>>n;for(int i=1; i<=n; i++) {cin>>a[i].x>>a[i].y;}sort(a+1,a+1+n,cmp);int cnt=0;int from=a[1].x;int base=a[1].y;for(int i=1; i<=n; i++) {if(a[i].x<=base) base=max(base,a[i].y);else {cnt+=(base-from);from=a[i].x;base=a[i].y;}}cnt+=(base-from);cout<<cnt<<endl;return 0;
}/*
in:
3
-1 1
5 11
2 9
out:
11
*/

三、AcWing 905：区间选点
（1）题目来源：https://www.acwing.com/problem/content/907/
（2）算法代码：https://blog.csdn.net/hnjzsyjyj/article/details/142840548

#include <bits/stdc++.h>
using namespace std;const int inf=0x3f3f3f3f;
const int maxn=1e5+5;struct Scope {int le,ri;
} a[maxn];bool up(Scope u,Scope v) {return u.ri<v.ri;
}int main() {int n;cin>>n;for(int i=0; i<n; i++) {cin>>a[i].le>>a[i].ri;}sort(a,a+n,up);int ans=0;int t_ri=-inf;for(int i=0; i<n; i++)if(t_ri<a[i].le) {ans++;t_ri=a[i].ri;}cout<<ans<<endl;return 0;
}/*
in:
3
-1 1
2 4
3 5
out:
2
*/

四、AcWing 906：区间分组
（1）题目来源：https://www.acwing.com/problem/content/908/
（2）算法代码：https://blog.csdn.net/hnjzsyjyj/article/details/145081557

#include <bits/stdc++.h>
using namespace std;const int maxn=1e5+5;
pair<int,int> a[maxn];
int n;int main() {cin>>n;for(int i=0; i<n; i++) {cin>>a[i].first>>a[i].second;}sort(a,a+n);priority_queue<int,vector<int>,greater<int>> q;for(int i=0; i<n; i++) {if(q.size() && a[i].first>q.top()) q.pop();q.push(a[i].second);}cout<<q.size()<<endl;return 0;
}/*
in:
10
-15 -8
-20 23
-2 11
2 22
18 23
11 27
-8 21
-18 14
-17 -12
-23 8out:
6
*/

五、AcWing 907：区间覆盖
（1）题目来源：https://www.acwing.com/problem/content/909/
（2）算法代码：https://blog.csdn.net/hnjzsyjyj/article/details/145083747

#include <bits/stdc++.h>
using namespace std;const int maxn=1e5+5;
struct Section {int x;int y;
} a[maxn];bool cmp(Section u,Section v) {if(u.x==v.x) return u.y<v.y;return u.x<v.x;
}int main() {int st,en;cin>>st>>en;int n;cin>>n;for(int i=1; i<=n; i++) {cin>>a[i].x>>a[i].y;}sort(a+1,a+1+n,cmp);int ans=0;bool flag=false;for(int i=1; i<=n; i++) {int j=i,t=INT_MIN;while(j<=n && a[j].x<=st) {t=max(t,a[j].y);j++;}if(t<st) break;ans++;if(t>=en) {flag=true;break;}st=t,i=j-1;}if(!flag) ans=-1;cout<<ans<<endl;return 0;
}/*
in:
1 5
3
-1 3
2 4
3 5out:
2
*/

六、AcWing 802：区间和
（1）题目来源：https://www.acwing.com/problem/content/804/
（2）算法代码：https://blog.csdn.net/hnjzsyjyj/article/details/143309807

#include <bits/stdc++.h>
using namespace std;typedef pair<int,int> PII;
const int maxn=3e5+5;
int a[maxn],s[maxn];
int n,m;
vector<int> alls;
vector<PII> add,query;int find(int x) { //discretizationint le=0;int ri=alls.size()-1;while(le<ri) {int mid=(le+ri)>>1;if(alls[mid]>=x) ri=mid;else le=mid+1;}return ri+1;
}int main() {cin>>n>>m;for(int i=0; i<n; i++) {int x,c;cin>>x>>c;alls.push_back(x);add.push_back({x,c});}for(int i=0; i<m; i++) {int le,ri;cin>>le>>ri;alls.push_back(le), alls.push_back(ri);query.push_back({le,ri});}//de-weightsort(alls.begin(),alls.end());alls.erase(unique(alls.begin(),alls.end()), alls.end());for(auto v:add) {int x=find(v.first);a[x]+=v.second;}for(int i=1; i<=alls.size(); i++) {s[i]=s[i-1]+a[i]; //prefix sum}for(auto v:query) {int le=find(v.first);int ri=find(v.second);cout<<s[ri]-s[le-1]<<endl;}return 0;
}/*
in:
3 3
1 2
3 6
7 5
1 3
4 6
7 8
out:
8
0
5
*/