Description
Igor found out discounts in a shop and decided to buy n items. Discounts at the store will last for a week and Igor knows about each item that its price now is ai, and after a week of discounts its price will be bi.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least k of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all n items.
In the first line there are two positive integer numbers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers b1, b2, ..., bn (1 ≤ bi ≤ 104) — prices of items after discounts (i.e. after a week).
Print the minimal amount of money Igor will spend to buy all n items. Remember, he should buy at least k items right now.
3 1
5 4 6
3 1 5
10
5 3
3 4 7 10 3
4 5 5 12 5
25
In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25.
题意：给我们当前的价格和一周后的价格，当前至少买k个物品，下周买n-k个物品，问总的花费最小
解法：以为是dp，后来用贪心过了，我们先比较两个价格的变化，优先买价格变化小的（只能买当前价格），接下来买价格中最便宜的。
我们记录价格变化，从小到大排序，然后前k个数字相加，然后买价格最便宜的
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <string.h> 4 #include <math.h> 5 #include <iostream> // C++头文件，C++完全兼容C 6 #include <algorithm> // C++头文件，C++完全兼容C 7 #define fre freopen("in.txt","r",stdin) //以文件代替控制台输入，比赛时很常用，能缩短输入测试样例的时间 8 #define INF 0x3f3f3f3f 9 #define inf 1e60 10 using namespace std; // C++头文件，C++完全兼容C 11 #define N 200005 // 宏定义 12 #define LL long long //宏定义 13 14 int a[N],b[N]; 15 struct P 16 { 17 int x,y,ans; 18 }H[N]; 19 int ans; 20 bool cmd(P a,P b) 21 { 22 return a.ans<b.ans; 23 } 24 25 int main() 26 { 27 int n,k; 28 cin>>n>>k; 29 for(int i=1;i<=n;i++) 30 { 31 cin>>H[i].x; 32 } 33 for(int i=1;i<=n;i++) 34 { 35 cin>>H[i].y; 36 H[i].ans=H[i].x-H[i].y; 37 } 38 sort(H+1,H+1+n,cmd); 39 for(int i=1;i<=k;i++) 40 { 41 ans+=H[i].x; 42 } 43 44 for(int i=k+1;i<=n;i++) 45 { 46 ans+=min(H[i].x,H[i].y); 47 } 48 cout<<ans<<endl; 49 return 0; 50 }