0
点赞
收藏
分享

微信扫一扫

B. Lorry (贪心)


​​点击打开链接​​

​​http://codeforces.com/contest/3/problem/B​​

B. Lorry


Description

A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).

Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body.

Input

The first line contains a pair of integer numbers n and v (1 ≤ n ≤ 105; 1 ≤ v ≤ 109), where n is the number of waterborne vehicles in the boat depot, and v is the truck body volume of the lorry in cubic metres. The following n lines contain the information about the waterborne vehicles, that is a pair of numbers ti, pi (1 ≤ ti; 1 ≤ pi ≤ 104), where ti is the vehicle type (1 – a kayak, 2 – a catamaran), and pi

Output

In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them.

Examples

input

3 2
1 2
2 7
1 3

output

7
2

题意:

一辆车可以承载体积V的货物,A种物品1个单位体积,B种2个单位体积,某种物品虽然体积相同但是贡献不同


给出N个物品它的物品类型(1,2)和贡献值。求这辆车最大可以拿到多大的贡献。

题解:贪心题啊。

明显地,对于体积相同的,我们肯定要选贡献最大的。那我们就从大到小选。我们先从大到小分别求这两种物品的前缀和。一起考虑两种物品,如果组成体积相同时,就选最大贡献的,如果贡献相同,就选体积最小的。


AC代码:


#include<bits/stdc++.h>
using namespace std;
int t1,t2;
int s1[100010],s2[100010];
pair<int,int>c1[100010],c2[100010];
bool cmp(pair<int,int>a,pair<int,int>b)
{
return a.first > b. first;
}
int main()
{
// freopen("in.txt","r",stdin);
int n,v;
cin>>n>>v;
for(int i=0;i<n;i++)
{
int t,p;
cin>>t>>p;
if(t==1){
c1[t1++] = make_pair(p,i+1);
}
else
c2[t2++] =make_pair(p,i+1);
}
sort(c1,c1+t1,cmp);
sort(c2,c2+t2,cmp);
s1[0] = s2[0] = 0;

for(int i=0;i<t1;i++){
s1[i+1]=s1[i]+c1[i].first;
}
for(int i=0;i<t2;i++){
s2[i+1]=s2[i]+c2[i].first;
}

int ans = -1;
int tmp = -1;
for(int i=0;i<=t2;i++)
{
if(i * 2 <= v && s1[ min(v-i*2, t1) ] + s2[i] > ans){
ans = s1[ min(v-i*2,t1) ] + s2[i] ;
tmp = i;
}
}
cout<<ans<<endl;
for(int i=0;i<min(v-tmp*2,t1);i++) cout<<c1[i].second<<endl;
for(int i=0;i<tmp;i++) cout<<c2[i].second<<endl;
return 0;
}




举报

相关推荐

0 条评论