华农校赛--G,用set比较大小，缩短时间复杂度

Array C

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 581  Solved: 101
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Description

 

   Giving two integers  and  and two arrays  and  both with length , you should construct an array  also with length  which satisfied:

 

1.0≤Ci≤Ai(1≤i≤n)

2.

 

and make the value S be minimum. The value S is defined as:

Input

 

   There are multiple test cases. In each test case, the first line contains two integers n(1≤n≤1000) andm(1≤m≤100000). Then two lines followed, each line contains n integers separated by spaces, indicating the array Aand B in order. You can assume that 1≤Ai≤100 and 1≤Bi≤10000 for each i from 1 to n, and there must be at least one solution for array C. The input will end by EOF.

Output

 

    For each test case, output the minimum value S as the answer in one line.

Sample Input

3 4 
2 3 4
1 1 1

Sample Output

6

HINT

 

    In the sample, you can construct an array [1,1,2](of course [1,2,1] and [2,1,1] are also correct), and  is 6.

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<cmath>
#include<algorithm>
#include<set>
using namespace std;
struct Node
{
    long long a;
    long long b;
    long long c;
    long long num;
    int i;
    bool operator < (const Node& t)const
    {
        return ((num>t.num)|| (num==t.num&&a<t.a)|| (num==t.num&&a==t.a&&b<t.b)||(num==t.num&&a==t.a&&b==t.b&&c<t.c)||(num==t.num&&a==t.a&&b==t.b&&c==t.c&&i<t.i));
    }

} node[1005];
set<Node>s;


int main()
{
    int n,m;

    while(scanf("%d%d",&n,&m)!=EOF)
    {
        long long res=0;
        long long sum=0;
        s.clear();
        for(int i=0; i<n; i++)
            scanf("%I64d",&node[i].a);
        for(int i=0; i<n; i++)
            scanf("%I64d",&node[i].b);
        for(int i=0; i<n; i++)
        {
            node[i].i=i;
            node[i].c=node[i].a;
            node[i].num=(2*node[i].c-1)*node[i].b;
            res+=node[i].c*node[i].c*node[i].b;
            sum+=node[i].a;
            s.insert(node[i]);
        }
        // cout<<res<<endl;
        Node tmp;
        set<Node>::iterator iter;
        for(int i=sum; i>m; i--)
        {
           // for(iter=s.begin(); iter!=s.end(); iter++)
               // cout<<iter->num<<"  ";
            tmp=(*s.begin());
            //cout<<tmp.num<<"***"<<res<<endl;

            s.erase(tmp);
            res-=tmp.num;
            tmp.c-=1;
            //out<<tmp.a<<endl;
            tmp.num=(2*tmp.c-1)*tmp.b;
            s.insert(tmp);

            //cout<<endl;

        }
        printf("%lld\n",res);

    }
    return 0;
}