is the order a rabbit （模拟+预处理+一点点贪心）

In a remote place, there is a rabbits market where rabbits can be bought and sold in every day. The trading rules in this market are as follows:

1.The market is available only in morning and afternoon in one day. And the rabbit transaction price may be different every day even different between morning and afternoon in a day.

2.It is prohibited that purchasing more than once in a day. And only one rabbit can be purchased at a time.

3. You can only sell once in a day, but you can sell rabbits unlimited quantities (provided that you have so many rabbits)

Mr.Cocktail has traded in this market for N days, suppose he has unlimited money to buy rabbits. Before the first day and after the last day, he has no rabbits. Now, in these N days, how much money did he earn the most?

Input
The first line contains a positive integer N, which means there are N days in total (1 \leq n \leq 10^5)(1≤n≤10 
5
 )

Next, there are N lines, each line contains two positive integers, representing the rabbit price in the morning and afternoon of the day a_i,b_i(1 \leq a_i ,b_i \leq 10^9)a 
i
​
 ,b 
i
​
 (1≤a 
i
​
 ,b 
i
​
 ≤10 
9
 ).

Output
Output an integer on a line to indicate the answer.

Sample 1
Inputcopy    Outputcopy
3
1 6
2 3
7 1
11
Sample 2
Inputcopy    Outputcopy
2
5 4
3 2
0
Note
In example 1, a rabbit was purchased for 1 in the morning of the first day, and a rabbit was purchased for1inthemorningofthefirstday,andarabbitwaspurchasedfor2 in the afternoon of the second day. On the morning of the third day, the two rabbits were sold, a total profit of $11.

The price of the rabbit in sample 2 continues to decrease, and it is impossible to make money through buying and selling, so choose not to make any buying and selling, and output the answer 0.

swjtu—春季集训 - Virtual Judge (vjudge.net)

思路：

• 按照时间顺序模拟，当后面的值没有比当前大，就全部卖出（贪心）
• 通过排序+树状数组解决是否有后面的值比自己大的情况

#include <bits/stdc++.h>
using namespace std;
#define ri register int
#define M 100005

template <class G> void read(G &x)
{
    x=0;int f=0;char ch=getchar();
    while(ch<'0'||ch>'9'){f|=ch=='-';ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    x=f?-x:x;
    return ;
}

int n,m;

struct dian{
    int val,id;
    bool operator <(const dian &t)const
    {
        return val>t.val;
    }
}p[M*2];
int val[M*2];
int flag[M*2];
long long num[M*2];
int cur=0;
void add(int a)
{
    while(a<=cur)
    {
        val[a]++;
        a+=a&(-a);
    }
}
int qu(int a)
{
    int ans=0;
    while(a>0)
    {
       ans+=val[a];
       a-=a&(-a);
    }
    return ans;
}
int main(){
    
    
    read(n);

    for(ri i=1;i<=n;i++)
    {
        read(p[++cur].val);p[cur].id=cur;
        num[cur]=p[cur].val;
        read(p[++cur].val);p[cur].id=cur;
        num[cur]=p[cur].val;
    }
    sort(p+1,p+1+cur);
    for(ri i=1;i<=cur;i++)
    {
        if(qu(cur)-qu(p[i].id-1)) 
        {
            flag[p[i].id]=1;
        }
        add(p[i].id);
    }
    long long ans=0;
    long long tmp=0;
    for(ri i=1;i<=cur;i+=2)
    {
        if(!flag[i])
        {
            ans+=tmp*num[i];tmp=0;
            if(flag[i+1]) ans-=num[i+1],tmp++;
            continue;
        }
        if(!flag[i+1])
        {
            ans-=num[i];tmp++;
            ans+=tmp*num[i+1];
            tmp=0;
            continue;
        }
        ans-=min(num[i],num[i+1]);
        tmp++;
    }
    printf("%lld",ans);
    return 0;
    
    
}

后记：

• 注意数组范围是2倍，循环的时候也是 2n，不要习惯写 成n了