Atcoder D - Knapsack 1 （背包）

D - Knapsack 1

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 100100 points

Problem Statement

There are NN items, numbered 1,2,…,N1,2,…,N. For each ii (1≤i≤N1≤i≤N), Item ii has a weight of wiwi and a value of vivi.

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

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

Constraints

• All values in input are integers.
• 1≤N≤1001≤N≤100
• 1≤W≤1051≤W≤105
• 1≤wi≤W1≤wi≤W
• 1≤vi≤1091≤vi≤109

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Input

Input is given from Standard Input in the following format:

NN WW
w1w1 v1v1
w2w2 v2v2
::
wNwN 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

Copy

90

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

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Sample Input 2 Copy

Copy

5 5
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000

Sample Output 2 Copy

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5000000000

The answer may not fit into a 32-bit integer type.

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

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

　　题意：有N个团队，每一个团队有w[i]个人，并且有v[i]的贡献值。

现在你有一个W的容量的公司，你需要从这n个团队中挑选出几个（每一个团队只能挑选一次），使之团队人数的总和小于W，并且贡献值的总和尽可能的大。

思路：裸的普通的背包问题。

我的AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#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 gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n,m;
int w[maxn];
ll v[maxn];
ll dp[maxn];
int main()
{
    gbtb;
    cin>>n>>m;
    repd(i,,n)
    {
        cin>>w[i]>>v[i];
    }
    repd(i,,n)
    {
        for(int j=m;j>=w[i];j--)
        {
            dp[j]=max(dp[j],1ll*dp[j-w[i]]+v[i]);
        }
    }
    cout<<dp[m];
    return ;
}

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