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—春季集訓 - Virtual Judge (vjudge.net)

思路：

• 就是拿到螞蟻爬的問題，碰撞相互變向，我們思考就讓他不變向，按著原來方向走就行了，
• 這樣 k秒過後，就是 答案的位置，
• 如何確定每一個位置具體是那個呢？
• 按照 原來的位置進行排序，因為不允許相互穿過去，所以k秒後，原來的排序大小還是不變

#include <bits/stdc++.h>
using namespace std;
#define ri register int
#define M  100055

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 ;
}

struct dian{
    int  val,id;
    int v;
    bool operator <(const dian &t)const
    {
        return val<t.val;
    }
}p[M];
int n,m;
int num[M];
long long ans[M];
int main(){
    
    read(n);read(m);
    for(ri i=1;i<=n;i++)
    {
        read(p[i].val);read(p[i].v);
        p[i].id=i;
    }
    sort(p+1,p+1+n);
    for(ri i=1;i<=n;i++)
    {
        num[i]=p[i].id;
    }
    for(ri i=1;i<=n;i++)
    {
        p[i].val+=p[i].v*m;
    }
    sort(p+1,p+1+n);
    for(ri i=1;i<=n;i++)
    {
        ans[num[i]]=p[i].val;
    }
    for(ri i=1;i<=n;i++)
    {
        if(i==1)
        printf("%lld",ans[i]); ////////////// ma de 
        else printf(" %lld",ans[i]);  
    }
    return 0;
    
}

後記：

• ！！！媽的，第一次遇到oj上不能允許某位有空格！！！
• 這個錯找了 25min ！！氣死了