B. Lorry(爆搜二分)

B. Lorry
time limit per test2 seconds
memory limit per test64 megabytes
inputstandard input
outputstandard 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 t i,  p i (1 ≤ t i ≤ 2; 1 ≤ p i ≤ 104), where t i is the vehicle type (1 – a kayak, 2 – a catamaran), and p i 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.

Examples
inputCopy
3 2
1 2
2 7
1 3
outputCopy
7
2

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题目大意：往车里塞体积1 或者体积2 的物件，每个物件有对应的价值。就是一个背包问题。不过就两种体积的物品，只是价值不一样。刚开始以为是贪心，越写越Wa。

思路：暴力搜索，枚举体积为1的物品，然后二分去找体积为2的物品塞背包。
坑点，最后输出加个空格，不加，疯狂WA。

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#include
#include
#include
#include
#include
#include
#include
#include
const int maxn=1e5+10;
const int INF=0x3f3f3f3f;
using namespace std;
struct node{
	int x,y;
}p1[maxn],p2[maxn];
bool cmp(node a,node b){
	return a.y>b.y;
}
int sum1[maxn],sum2[maxn];
int main(){
	int n,v,len1=0,len2=0;
	cin>>n>>v;
	int x,y,ans=0;
	for(int i=1;i<=n;i++){
		scanf("%d %d",&x,&y);
		if(x==1){
			p1[++len1].x = i; 
			p1[len1].y = y;
		}
		else{
			p2[++len2].x = i; 
			p2[len2].y = y;
		} 
	}
	sort(p1+1,p1+1+len1,cmp);
	sort(p2+1,p2+1+len2,cmp);
	for(int i=1;i<=len1;i++){
		sum1[i] = sum1[i-1] + p1[i].y;
	}
	for(int i=1;i<=len2;i++){
		sum2[i] = sum2[i-1] + p2[i].y;
	}
	x=y=0;
	for(int i=0;i<=len1;i++){
		if(i>v) break;
		int l=0,r=len2,j=0;
		while(l<=r){
			int mid = (l+r)/2;
			if(mid*2<=(v-i)) j=mid,l=mid+1;
			else r=mid-1;
		}
		int tmp = sum1[i] + sum2[j];
			if(tmp>ans){
				ans = tmp;	
				x = i,y = j;  
			}	
	}
	cout<<ans<<endl;
	for(int i=1;i<=x;i++) printf("%d ",p1[i].x);
	for(int i=1;i<=y;i++) printf("%d ",p2[i].x);
}