Atcoder E - Knapsack 2 （01背包进阶版 ex ）

 

E - Knapsack 2

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 100 points

Problem Statement

There are vi.

Taro has decided to choose some of the W.

Find the maximum possible sum of the values of items that Taro takes home.

Constraints

• All values in input are integers.
• 1≤N≤100
• 1≤W≤109
• 1≤wi≤W
• 1≤vi≤103

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Input

Input is given from Standard Input in the following format:

W
v1
v2
:
vN

Output

Print the maximum possible sum of the values of items that Taro takes home.

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Sample Input 1 Copy

Copy

3 8
3 30
4 50
5 60

Sample Output 1 Copy

Copy

90

Items 30+60=90.

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Sample Input 2 Copy

Copy

1 1000000000
1000000000 10

Sample Output 2 Copy

Copy

10

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Sample Input 3 Copy

Copy

6 15
6 5
5 6
6 4
6 6
3 5
7 2

Sample Output 3 Copy

Copy

17

Items 6+6+5=17.

题目链接：https://atcoder.jp/contests/dp/tasks/dp_e

思路：体积虽然很huge，但价值很小。把最大化价值，转成最小化体积，就还是原来的01背包了。（RUSH_D_CAT大佬指点的）

很不错的一个背包优化的思路，细节见我的代码。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll dp[maxn];
ll n,W;
ll v[maxn];
ll w[maxn];
int main()
{
    memset(dp,0x3f,sizeof(dp));
    gbtb;
    cin>>n>>W;
    repd(i,1,n)
    {
        cin>>w[i]>>v[i];
    }
    ll ans=0ll;
    dp[0]=0;
    repd(i,1,n)
    {
        for(int j=1e5;j>=v[i];--j)
        {
            {
                dp[j]=min(dp[j],dp[j-v[i]]+w[i]);
                if(dp[j]<=W)
                    ans=max(ans,(ll)(j));
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}