E - Knapsack 2

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

Copy

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

Copy

1 1000000000
1000000000 10

Sample Output 2 Copy

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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

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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 $\mathrm{6+6+5=176+6+5=17}$.

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod =1e9+7;
const int N = 1e5+5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}
bool prime(int x) {
    if (x < 2) return false;
    for (int i = 2; i * i <= x; ++i) {
        if (x % i == 0) return false;
    }
    return true;
}
ll qpow(ll m, ll k, ll mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}      

int main()
{
    ll n,w;
    cin >> n >> w;
    vector<pair<ll, ll>> a(n + 1);
    vector<vector<ll>> dp(n + 1,vector<ll>(N));
    rep(i, 1, n)
    {
        cin >> a[i].first>>a[i].second;
    }
    rep(i, 1, 1e5)
    {
        dp[0][i] = inf;
    }
    rep(i, 1, n)
    {
        rep(j, 1, 1e5)
        {
            if ( j >=a[i].second)
                dp[i][j] = min(dp[i - 1][j],dp[i - 1][j-a[i].second] + a[i].first);
            else
                dp[i][j] = dp[i - 1][j];
        }
    }
    ll res = 0;
    dec(i,1e5,0)
    {
        if (dp[n][i] <= w)
        {
            res = i;
            break;
        }
    }
    cout << res << endl;
    return 0;
}