CF3B 贪心

http://codeforces.com/problemset/problem/3/B

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 test(s)

input

3 21 22 71 3

output

72

/**CF3B 贪心题目大意；给定一个指定容量的背包，有n种物品，每种物品给定体积(1 or 2)和价值，问背包最多能容的价值是多少，并且求出装了哪几个物品解题思路：背包问题，可惜背包容量过大。采用贪心的思想来做：按照价值/体积的递减顺序排序，从头依次装物品，直到装不下跳出循环。          此时如果背包没有装满，那么肯定差1，有两种方案：1，从后面加价值最大的一个体积为1的物品。2，从前面取价值最小体积为1的物品去掉          然后在后面加上价值最大的一个体积为2的物品，二者取最优*/#include <string.h>#include <algorithm>#include <iostream>#include <stdio.h>using namespace std;struct note{    int x,y;    int id,judge;    bool operator < (const note & other)const    {        return x*other.y<other.x*y;    }} a[1000505];int n,m,b[1050005];int main(){    while(~scanf("%d%d",&n,&m))    {        for(int i=0; i<n; i++)        {            scanf("%d%d",&a[i].x,&a[i].y);            a[i].id=i+1;            a[i].judge=0;        }        sort(a,a+n);        int sum=0,flag=-1;        for(int i=0; i<n; i++)        {            if(m-a[i].x>=0)            {                sum+=a[i].y;                a[i].judge=1;                m-=a[i].x;                flag=i;            }            else            {                break;            }        }        if(m>0&&flag<n)        {            int sum1=-1,sum2=-1,flag1,flag2=-1,flag3;            ///后加1            for(int i=flag+1; i<n; i++)            {                if(a[i].x==1)                {                    sum1=sum+a[i].y;                    flag1=i;                    break;                }            }            ///前去1后加2            for(int i=flag; i>=0; i--)            {                if(a[i].x==1)                {                    sum2=sum-a[i].y;                    flag2=i;                    for(int j=flag+1; j<n; j++)                    {                        if(a[j].x==2)                        {                            sum2+=a[j].y;                            flag3=j;                            break;                        }                    }                    break;                }            }            if(sum1>=sum2&&sum1>sum)            {                sum=sum1;                a[flag1].judge=1;            }            else if(sum2>=sum1&&sum2>sum)            {                sum=sum2;                a[flag2].judge=0;                a[flag3].judge=1;            }        }        printf("%d\n",sum);        int num=0;        for(int i=0; i<n; i++)        {            if(a[i].judge==1)            {                b[num++]=a[i].id;            }        }        for(int i=0; i<num; i++)        {            printf(i==num-1?"%d\n":"%d ",b[i]);        }    }    return 0;}/**3 21 22 71 33 21 21 31 43 32 82 71 33 32 82 72 33 31 41 32 5*/