首页 > 技术文章 > JZOJ 5849 d

traveller-ly 2018-08-25 23:28 原文

Description

Input

Output

Data Constraint

 

 

做法:考虑贪心使得mina*minb最大,先以a为关键字排序,然后枚举删除最小的0~m个ai,对应删除最小的bi,然后更新答案。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <queue>
 5 #include <algorithm>
 6 #define M 100007
 7 #define LL long long
 8 #define rep(i,a,b) for(int i=a;i<=b;i++)
 9 using namespace std;
10 int T,n,m;
11 LL ans;
12 struct arr{
13     int a,b,site;
14 }f[M];
15 bool v[M];
16 
17 LL max(LL a,LL b) {return a>b?a:b;}
18 
19 int Cmp(arr x,arr y){
20     return x.a<y.a;
21 }
22 
23 bool operator <(const arr &x,const arr &y){
24     return x.b>y.b;
25 }
26 
27 void Init(){
28     memset(f,0,sizeof(f));
29     scanf("%d%d",&n,&m);
30     rep(i,1,n)    scanf("%d%d",&f[i].a,&f[i].b),f[i].site=i;
31     sort(1+f,f+n+1,Cmp);
32 }
33 
34 void Work(){
35     priority_queue<arr> q;
36     ans=0;
37     rep(i,m+1,n) q.push(f[i]);
38     arr now=q.top();
39     ans=max(ans,(LL)f[m+1].a*now.b); 
40     for(int i=m;i>=1;i--){
41         q.pop();
42         q.push(f[i]);
43         now=q.top();
44         ans=max(ans,(LL)f[i].a*now.b);
45     }
46     now=q.top();
47     ans=max(ans,(LL)f[1].a*now.b);
48     printf("%lld\n",ans);
49 }
50 
51 int main() {
52     freopen("d.in","r",stdin);
53     freopen("d.out","w",stdout);
54     scanf("%d",&T);
55     for(;T--;){
56         Init();
57         Work();
58     }
59 }
View Code

 

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