Codeforces Round #415 B Summer sell-off

Summer sell-off
time limit per test1 second
memory limit per test256 megabytes

Description
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.

Shop, where Noora is working, has a plan on the following n days. For each day sales manager knows exactly, that in i-th day ki products will be put up for sale and exactly li clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren’t any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren’t kept on the next day and are sent to the dump.

For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any f days from n next for sell-outs. On each of f chosen days the number of products were put up for sale would be doubled. Thus, if on i-th day shop planned to put up for sale ki products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·ki products. Consequently, there is an opportunity to sell two times more products on days of sell-out.

Noora’s task is to choose f days to maximize total number of sold products. She asks you to help her with such a difficult problem.

Input
The first line contains two integers n and f (1 ≤ n ≤ 105, 0 ≤ f ≤ n) denoting the number of days in shop’s plan and the number of days that Noora has to choose for sell-out.

Each line of the following n subsequent lines contains two integers ki, li (0 ≤ ki, li ≤ 109) denoting the number of products on the shelves of the shop on the i-th day and the number of clients that will come to the shop on i-th day.

Output
Print a single integer denoting the maximal number of products that shop can sell.

Examples
Input
4 2
2 1
3 5
2 3
1 5
Output
10
Input
4 1
0 2
0 3
3 5
0 6
Output
5

Note
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.

In the second example it is possible to sell 5 products, if you choose third day for sell-out.

题目大意：这题大概就是在n组库存与销售量关系中，找出f组，使其库存翻倍从而提高销售额，我们的目的就是找出利益最大化的f组。
看上去挺水的，我一开始直接开三个数组想暴力排序+查找，果不其然，好像是第12组数据就WA了(/”≡ _ ≡)/~┴┴
唉，之后就老老实实结构体数组排序呗，顺带写了个cmp，快速sort一下就好了，感觉写的比暴力还要快？？？

#include<bits/stdc++.h>
using namespace std;
int n,m;
typedef long long ll;
struct edge
{
    ll a,b,c;
}d[100005];
 bool cmp(edge A,edge B)
{
    return A.c<B.c;
}
int mak[100005];
int main()
{
    memset(mak,0,sizeof(mak));
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
    {
        scanf("%lld%lld",&d[i].a,&d[i].b);
        long long int num=d[i].a*2;
        if(d[i].a<=d[i].b)//只有这种情况能提高销售额
        {
            if(num>=d[i].b)
                d[i].c=d[i].b-d[i].a;
            else if(num<=d[i].b)
                d[i].c=num-d[i].a;
            //mak[i]=1;
            //printf("%d\n",i);
        }
    }
    sort(d,d+n,cmp);
    long long int sum=0;
    for(int i=n-1;i>=0;i--)
    {
        if(i>=n-m)
        {
            if(d[i].a*2>=d[i].b)
                sum+=d[i].b;
            else
                sum+=d[i].a*2;
        }
        else
            sum+=min(d[i].a,d[i].b);
    }
    printf("%lld\n",sum);
}