题目大意是有体积为v的背包，有体积为1和2的两种物品若干，这些物品都有各自的价值。求如何取这些物品可使背包中物品的价值最大。

开始一看到是背包就傻眼了==因为数据量太大1 ≤ n ≤ 105; 1 ≤ v ≤ 109 

搜题解有说用优先队列做的，好麻烦==有一种思路我很喜欢：既然涉及到背包或者贪心这类DP问题，一定是取最优的解。即如果把两种船只分别存到两个数组里，再由大到小排序，最终的结果一定是两个数组各自的前1~M个，且无跳跃，分别都是连续的。那么假设我遍历大船的情况从0到船只数，意味着分别是有0个大船、1个大船、、、所有大船，剩余的费用给小船。在循环中记录每次的最大值及大船、小船的个数。

还有就是最后输出结果中第二行表示船的序号，另：G++ WA了 VC++ AC了

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 ≤ 2; 1 ≤ pi ≤ 104), where ti is the vehicle type (1 – a kayak, 2 – a catamaran), and pi is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file.

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.

Sample Input

Input 
3 2
1 2
2 7
1 3
Output 
7
2
#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
struct ship
{
    int id,cost;
}ka[100000],ca[100000];
bool cmp(ship a,ship b)
{
    return a.cost>b.cost;
}
int total1[100000],total2[100000];
int n,v,sum[100000],t;
int main()
{
  // freopen("cin.txt","r",stdin);
    //freopen("cout.txt","w",stdout);
    while(~scanf("%d%d",&n,&v))
    {
        int num1=0,num2=0;
        for(int i=1;i<=n;i++) {
            scanf("%d",&t);
            if(t==1) {scanf("%d",&ka[num1].cost);ka[num1++].id=i;}
            else {scanf("%d",&ca[num2].cost);ca[num2++].id=i;}
        }
        sort(ka,ka+num1,cmp);
        //for(int i=0;i<num1;i++) printf("i=%d %d  ",ka[i].id,ka[i].cost);printf("\n");
        sort(ca,ca+num2,cmp);
        //for(int i=0;i<num2;i++) printf("i=%d %d  ",ca[i].id,ca[i].cost);printf("\n");
        int p1=0,p2=0,ans=0;
        total1[0]=0,total2[0]=0,total1[1]=ka[0].cost,total2[1]=ca[0].cost;
        for(int i=1;i<num1;i++) total1[i+1]=total1[i]+ka[i].cost;
        for(int i=1;i<num2;i++) total2[i+1]=total2[i]+ca[i].cost;
        int cnt=0,t,tmp,max0=-1,max1=-1,max2=-1;
        for(int i=0;i<=num2;i++)
        {
            if(i*2>v) break;
            t=v-i*2;
            if(t>num1) t=num1;
            if(total2[i]+total1[t]>max0)
            {
                max0=total2[i]+total1[t];
                max1=t;
                max2=i;
            }
        }
        printf("%d\n",max0);
        for(int i=0;i<max1;i++) printf("%d ",ka[i].id);
        for(int i=0;i<max2;i++) printf("%d ",ca[i].id);
        printf("\n");
    }
    return 0;
}