Codeforces Gym 100513D D. Data Center 前缀和 排序

D. Data Center

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/560/problem/B

Description

The startup "Booble" has shown explosive growth and now it needs a new data center with the capacity of m petabytes. Booble can buy servers, there are n servers available for purchase: they have equal price but different capacities. The i-th server can store ai petabytes of data. Also they have different energy consumption — some servers are low voltage and other servers are not.

Booble wants to buy the minimum number of servers with the total capacity of at least m petabytes. If there are many ways to do it Booble wants to choose a way to maximize the number of low voltage servers. Booble doesn't care about exact total capacity, the only requirement is to make it at least m petabytes.

Input

The first line contains two integer numbers n and m (1 ≤ n ≤ 2·105, 1 ≤ m ≤ 2·1015) — the number of servers and the required total capacity.

The following n lines describe the servers, one server per line. The i-th line contains two integers ai, li (1 ≤ ai ≤ 1010, 0 ≤ li ≤ 1), where ai is the capacity, li = 1 if server is low voltage and li = 0 in the opposite case.

It is guaranteed that the sum of all ai is at least m

Output

Print two integers r and w on the first line — the minimum number of servers needed to satisfy the capacity requirement and maximum number of low voltage servers that can be bought in an optimal r servers set.

Print on the second line r distinct integers between 1 and n — the indices of servers to buy. You may print the indices in any order. If there are many solutions, print any of them.

Sample Input

4 10
3 1
7 0
5 1
4 1

Sample Output

2 1
4 2

HINT

题意

有个人要买电池，要求买尽量少的电池，使得满足容量大于等于m，并且使得低能耗的电池尽量多

题解：

先排个序，判断出得至少买多少个电池

然后开始暴力枚举低能耗的电池个数，肯定优先拿电量大的，然后扫一遍就好了

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
const int maxn=;
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//**************************************************************************************
 
struct node
{
    ll x,y,z;
};
bool cmp(node a,node b)
{
    return a.x>b.x;
}
bool cmp1(node a,node b)
{
    if(a.y==b.y)
        return a.x>b.x;
    return a.y>b.y;
}
node a[maxn];
ll sum1[maxn];
ll sum2[maxn];
int main()
{
    int n=read();
    ll m=read();
    for(int i=;i<=n;i++)
        a[i].x=read(),a[i].y=read(),a[i].z=i;
    sort(a+,a+n+,cmp);
    ll sum=;
    int num=;
    for(int i=;i<=n;i++)
    {
        num=i;
        sum+=a[i].x;
        if(sum>=m)
            break;
    }
    sort(a+,a+n+,cmp1);
    int flag=;
    for(int i=;i<=n;i++)
        if(a[i].y!=)
        {
            flag=i;
            break;
        }
    int num1=,num2=;
    if(flag==)
        flag=n+;
    for(int i=;i<flag;i++)
    {
        sum1[num1]=sum1[num1-]+a[i].x;
        num1++;
    }
    for(int i=flag;i<=n;i++)
    {
        sum2[num2]=sum2[num2-]+a[i].x;
        num2++;
    }
    for(int i=num;i>=;i--)
    {
        if(sum1[i]+sum2[num-i]>=m)
        {
            cout<<num<<" "<<i<<endl;
            for(int j=;j<=i;j++)
                cout<<a[j].z<<" ";
            for(int j=flag;j<flag+num-i;j++)
                cout<<a[j].z<<" ";
            cout<<endl;
            return ;
        }
    }
}