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 : $100100$ points

Problem Statement

There are $\mathrm{NN}$ items, numbered $\mathrm{1,2,\dots ,N1,2,\dots ,N}$. For each $\mathrm{ii}$ ($\mathrm{1\le i\le N1\le i\le N}$), Item $\mathrm{ii}$ has a weight of ${\mathrm{wiwi}}^{}$ and a value of ${\mathrm{vivi}}^{}$.

Taro has decided to choose some of the $\mathrm{NN}$ items and carry them home in a knapsack. The capacity of the knapsack is $\mathrm{WW}$, which means that the sum of the weights of items taken must be at most $\mathrm{WW}$.

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

Constraints

• All values in input are integers.
• $\mathrm{1\le N\le 1001\le N\le 100}$
• $\mathrm{1\le W\le 1091\le W\le 109}$
• $\mathrm{1\le wi\le W1\le wi\le W}$
• $\mathrm{1\le vi\le 1031\le vi\le 103}$

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Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$ $\mathrm{WW}$
${\mathrm{w1w1}}^{}$ ${\mathrm{v1v1}}^{}$
${\mathrm{w2w2}}^{}$ ${\mathrm{v2v2}}^{}$
$::$
${\mathrm{wNwN}}^{}$ ${\mathrm{vNvN}}^{}$

Output

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

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Sample Input 1 Copy

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3 8
3 30
4 50
5 60

Sample Output 1 Copy

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90

Items $11$ and $33$ should be taken. Then, the sum of the weights is $\mathrm{3+5=83+5=8}$, and the sum of the values is $\mathrm{30+60=9030+60=90}$.

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Sample Input 2 Copy

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1 1000000000
1000000000 10

Sample Output 2 Copy

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10

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Sample Input 3 Copy

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

Sample Output 3 Copy

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17

Items $\mathrm{2,42,4}$ and $55$ should be taken. Then, the sum of the weights is $\mathrm{5+6+3=145+6+3=14}$, and the sum of the values is $\mathrm{6+6+5=176+6+5=17}$.

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';
        }
    }
}