E. Selling Souvenirs
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

After several latest reforms many tourists are planning to visit Berland, and Berland people understood that it's an opportunity to earn money and changed their jobs to attract tourists. Petya, for example, left the IT corporation he had been working for and started to sell souvenirs at the market.

This morning, as usual, Petya will come to the market. Petya has n different souvenirs to sell; ith souvenir is characterised by its weight wi and cost ci. Petya knows that he might not be able to carry all the souvenirs to the market. So Petya wants to choose a subset of souvenirs such that its total weight is not greater than m, and total cost is maximum possible.

Help Petya to determine maximum possible total cost.

Input

The first line contains two integers n and m (1 ≤ n ≤ 100000, 1 ≤ m ≤ 300000) — the number of Petya's souvenirs and total weight that he can carry to the market.

Then n lines follow. ith line contains two integers wi and ci (1 ≤ wi ≤ 3, 1 ≤ ci ≤ 109) — the weight and the cost of ith souvenir.

Output

Print one number — maximum possible total cost of souvenirs that Petya can carry to the market.

Examples
input
Copy 
1 1
2 1
output 
0
input
Copy 
2 2
1 3
2 2
output 
3
input
Copy 
4 3
3 10
2 7
2 8
1 1
output 
10

题意：正常的01背包描述，不正常的数据，卡时间。

思路：

① 观察到数据中，w仅有123

②暴力枚举所有选择重量为3的情况，那么剩余背包容积为M=m-3*i

③假设有w=2的x个，w=1的y个。2x+3y<=M,求max(sum[2][x]+sum[1][(M-2x)/3])

④很明显，一旦x确定，y的最大值也可以确定。很明显存在最大值(等待证明)，凸字型三分求最值。

#include <bits/stdc++.h>
#define bug cout <<"bug"<<endl

using namespace std;
typedef long long ll;

const int MAX_N=1e5+3;
const int MOD=1e9+7;
vector <int> p[4];
ll sum[4][MAX_N];
ll n,m;

ll check(int n3,int n2){
    int n1=m-3*n3-2*n2;
    return sum[2][n2]+sum[1][min(n1,(int)p[1].size())];
}

int main(void){
    cin >> n>>m;
    for(int i=1;i<=n;i++){
        int num,temp;
        scanf("%d%d",&num,&temp);
        p[num].push_back(temp);
    }
    for(int i=1;i<=3;i++){
        sort(p[i].begin(),p[i].end(),greater<int>());
        for(int j=0;j<(int)p[i].size();j++){
            sum[i][j+1]=sum[i][j]+p[i][j];
//            cout << sum[i][j+1]<<endl;
        }
    }
    int cnt=min((ll)p[3].size(),m/3);
    ll ans=0;
    for(int i=0;i<=cnt;i++){
        int l=0,r=min((ll)p[2].size(),(m-3*i)/2);
        while(r>l+1){
            ll midl=(l+r)>>1;
            ll midr=(midl+r)>>1;
            if(check(i,midl)>check(i,midr)) r=midr;
            else    l=midl;
//            printf("midl=%d midr=%d\n",l,r);
        }
        ans=max(ans,max(check(i,l),check(i,r))+sum[3][i]);//check单调的情况
    }
    cout << ans << endl;
}