Codeforces 3B. Lorry

B. Lorry

time limit per test
2 seconds

memory limit per test
64 megabytes

input
standard input

output
standard output

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 2
1 2
2 7
1 3

output

7
2

题意：特殊的背包问题， 背包的容量是v， 有两张可以选择的物品，体积1的物品有x件， 体积2的物品有y件， 体积相同的物品的能力值并不相同， 问你该怎么装， 使的总能力值最大。

思路：贪心的思路，将体积1和2的物品分别按照能力值大小排序，  首先将体积是2的物品按序装到背包， 然后将体积是1的物品按序装到背包， 如果背包满了， 还有体积是1的剩余， 则将体积1的两个组合或者1个（只剩一个了）从后往前替换体积为2的物体（能力值要比体积为2的大才替换）

代码：

#include <stdio.h>
#include <algorithm>
#include <vector>
using namespace std;

const int MAXN = 100005;

struct Vehicle
{
int id, cap;
bool operator<(const Vehicle &v) const
{
return cap > v.cap;
}
Vehicle(){}
Vehicle(int x, int y) : id(x), cap(y){}
};

int cap[MAXN];
vector<Vehicle> veh1;
vector<Vehicle> veh2;

int main()
{
#ifdef _LOCAL
freopen("F://input.txt", "r", stdin);
#endif
int n, v;
scanf("%d%d", &n, &v);
for(int i = 1; i <= n; ++i)
{
int a, b;
scanf("%d%d", &a, &b);
cap[i] = b;
if(a == 1)
veh1.push_back(Vehicle(i, b));
else
veh2.push_back(Vehicle(i, b));
}
sort(veh1.begin(), veh1.end());
sort(veh2.begin(), veh2.end());
vector<int> ans;
int totalCap = 0;
for(int i = 0; i < veh2.size() && v >= 2; ++i, v -= 2)
{
totalCap += veh2[i].cap;
ans.push_back(veh2[i].id);
}
int endFlag = ans.size() - 1;
int startFlag = 0;
for( ; startFlag < veh1.size() && v; ++startFlag, --v)
{
ans.push_back(veh1[startFlag].id);
totalCap += veh1[startFlag].cap;
}
for(int i = endFlag, j = startFlag; i >= 0 && j < veh1.size(); --i, ++j)
{
if(j + 1 < veh1.size())
{
if(cap[ans[i]] >= veh1[j].cap + veh1[j + 1].cap)
break;
totalCap = totalCap - cap[ans[i]] + veh1[j].cap + veh1[j + 1].cap;
ans[i] = veh1[j].id;
ans.push_back(veh1[j + 1].id);
j = j + 1;
}
else  if(cap[ans[i]] < veh1[j].cap)
{
totalCap = totalCap - cap[ans[i]] + veh1[j].cap;
ans[i] = veh1[j].id;
}
}
printf("%d\n", totalCap);
for(int i = 0; i < ans.size(); ++i)
printf("%d ", ans[i]);
return 0;
}