题目大意：给你两种物品，每种物品有一个价值和花费，花费只有两种，一种花费为 ， 一种花费为2.、

给你一个背包容量为v, 求当前容量下所能达到的最大价值。

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#include <iostream>

#include <cmath>

#include <algorithm>

#include <string>

#include <cstring>

#include <cstdio>

#include <vector>

#include <cstdlib>

using namespace std;

typedef __int64 LL;

const LL INF = 0xffffff;

const int maxn = ;

const LL MOD = 1e9+;

struct node

{

    int cost;///花费

    int value;///价值

    int index;///下标

    bool friend operator < (node A, node B)

    {

        return A.value > B.value;///按照单价进行排序

    }

}One[maxn], Tow[maxn], P;

int sum[maxn] = {}, ans[maxn];

int main()

{

    int n, Cost;

    int num1 = , num2 = ;///Last 最后一个价值为 1 物品的下标。

    scanf("%d %d", &n, &Cost);

    for(int i=; i<=n; i++)

    {

        scanf("%d %d",&P.cost, &P.value);

        P.index = i;

        if(P.cost == )

            One[num1 ++] = P;

        else

            Tow[num2 ++] = P;

    }

    sort(One+, One + num1+);

    sort(Tow+, Tow + num2+);

    Tow[].cost = , Tow[].value = ;

    for(int i=; i< num1; i++)

        sum[i] += sum[i-] + One[i].value;

    int Index = , Max = , Temp, CurValue = , CurCost = ;

    for(int i=; i< num2; i++)

    {

        CurValue += Tow[i].value;

        CurCost += Tow[i].cost;

        if(CurCost > Cost)

            break;

        if(Cost - CurCost >= num1-)

            Temp =  CurValue + sum[num1-];

        else

            Temp =  CurValue + sum[Cost - CurCost];

        if(Temp > Max)

        {

            Max = Temp;

            Index = i;

        }

    }

    printf("%d\n", Max);

    int k = ;

    for(int i=; i<=Index; i++)

        ans[k ++] = Tow[i].index;

    for(int i=; i<=(Cost-Index*)&& i<num1; i++)

        ans[k ++] = One[i].index;

    for(int i=; i<k- ;i++)

        printf("%d ", ans[i]);

    if(k != )

        printf("%d\n", ans[k-]);

    return ;

}

/*

10 10

1 14

2 15

2 11

2 12

2 9

1 14

2 15

1 9

2 11

2 6

*/