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.

[b]Input[/b]

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

[b]Output[/b]

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.

[b]Sample Input[/b]

4 10
3 1
7 0
5 1
4 1

[b]Sample Output[/b]

2 1
4 2

HINT

[b]题意
[/b]

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

[b]题解：[/b]

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

[b]代码[/b]

#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=202501;
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';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=1;i<=n;i++)
a[i].x=read(),a[i].y=read(),a[i].z=i;
sort(a+1,a+n+1,cmp);
ll sum=0;
int num=0;
for(int i=1;i<=n;i++)
{
num=i;
sum+=a[i].x;
if(sum>=m)
break;
}
sort(a+1,a+n+1,cmp1);
int flag=0;
for(int i=1;i<=n;i++)
if(a[i].y!=1)
{
flag=i;
break;
}
int num1=1,num2=1;
if(flag==0)
flag=n+1;
for(int i=1;i<flag;i++)
{
sum1[num1]=sum1[num1-1]+a[i].x;
num1++;
}
for(int i=flag;i<=n;i++)
{
sum2[num2]=sum2[num2-1]+a[i].x;
num2++;
}
for(int i=num;i>=0;i--)
{
if(sum1[i]+sum2[num-i]>=m)
{
cout<<num<<" "<<i<<endl;
for(int j=1;j<=i;j++)
cout<<a[j].z<<" ";
for(int j=flag;j<flag+num-i;j++)
cout<<a[j].z<<" ";
cout<<endl;
return 0;
}
}
}