Acwing2024蓝桥杯区间合并

模板:

#define x first
#define y second
typedef pair<int, int>pii;
pii seg[N];
 
sort(seg,seg+n);
int l=seg[0].x,r=seg[0].y;
for (int i=1;i<n;i++){
    if (seg[i].x<=r) r=max(r,seg[i].y);
    else l=seg[i].x,r=seg[i].y;
}

题目:

AcWing 1343. 挤牛奶

#include<iostream>
#include<algorithm>
#define x first
#define y second
using namespace std;
const int N=5005;
typedef pair<int,int>pii;
pii seg[N];
int n,ans1,ans2;
int main(){
    cin>>n;
    for(int i=0;i<n;i++) cin>>seg[i].x>>seg[i].y;
    sort(seg,seg+n);
    int l=seg[0].x,r=seg[0].y;
    for(int i=0;i<n;i++){
        if(seg[i].x<=r){
            r=max(r,seg[i].y);
            ans1=max(ans1,r-l);
        }
        else{
            ans2=max(ans2,seg[i].x-r);
            l=seg[i].x,r=seg[i].y;
            ans1=max(ans1,r-l);
        }
    }
    cout<<ans1<<" "<<ans2<<endl;
    return 0;
}

AcWing 803. 区间合并

#include<iostream>
#include<algorithm>
#define x first
#define y second
using namespace std;
const int N=1e5+5;
typedef pair<int,int>pii;
pii seg[N];
long long ans=1;
int main(){
    int n;cin>>n;
    for(int i=0;i<n;i++) cin>>seg[i].x>>seg[i].y;
    sort(seg,seg+n);
    int l=seg[0].x,r=seg[0].y;
    for(int i=0;i<n;i++){
        if(seg[i].x<=r) r=max(r,seg[i].y);
        else l=seg[i].x,r=seg[i].y,ans++;
    }
    cout<<ans<<endl;
    return 0;
}

AcWing 422. 校门外的树

#include<iostream>
#include<algorithm>
#define x first
#define y second
using namespace std;
const int N=105;
typedef pair<int,int>pii;
pii seg[N];
int ans;
int main(){
    int len,n;cin>>len>>n;
    for(int i=0;i<n;i++) cin>>seg[i].x>>seg[i].y;
    sort(seg,seg+n);
    int l=seg[0].x,r=seg[0].y;
    for(int i=0;i<n;i++){
        if(seg[i].x<=r) r=max(r,seg[i].y);
        else ans+=r-l+1,l=seg[i].x,r=seg[i].y;
    }
    ans+=r-l+1;
    cout<<len+1-ans<<endl;
    return 0;
}

AcWing 5407. 管道(第十四届省赛)

#include<iostream>
#include<algorithm>
#define x first
#define y second
using namespace std;
const int N=1e5+5;
int n,len,id[N],timee[N];
typedef pair<int,int>pii;
pii seg[N];
bool check(long long value){
    int j=0,flag=1;
    //划分区间
    for(int i=1;i<=n;i++){
        if(timee[i]<=value){
            //flag进行标记
            flag=0;
            int temp=value-timee[i];
            seg[j].x=max(1,id[i]-temp),seg[j].y=min(len,id[i]+temp),j++;
        }
    }
    //如果flag==1则该时间绝对不可能
    if(flag) return false;
    //排序
    sort(seg,seg+j);
    //合并区间
    int l=seg[0].x,r=seg[0].y,cnt=0;
    for(int i=0;i<j;i++){
        if(seg[i].x<=r+1) r=max(r,seg[i].y);
        else{
            //cnt计数
            cnt+=r-l+1;
            l=seg[i].x;
            r=seg[i].y;
        }
        //最后一次合并区间后,cnt计数
        if(i==j-1) cnt+=r-l+1;
    }
    //check判断bool
    if(cnt==len) return true;
    else return false;
}
int main(){
    //输入
    cin>>n>>len;
    for(int i=1;i<=n;i++) cin>>id[i]>>timee[i];
    //二分查找
    int left=1,right=2e9;//注意right的值为2e9
    while(left<right){
        //注意mid开浪浪,否则最大测试数据会爆
        long long mid=(long long)left+right>>1;
        if(check(mid)) right=mid;
        else left=mid+1;
    }
    //输出
    cout<<left<<endl;
    return 0;
}