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题目链接

题目

题目描述

Farmer John's cows love their video games! FJ noticed that after playing these games that his cows produced much more milk than usual, surely because contented cows make more milk.
The cows disagree, though, on which is the best game console. One cow wanted to buy the Xbox 360 to play Halo 3; another wanted to buy the Nintendo Wii to play Super Smash Brothers Brawl; a third wanted to play Metal Gear Solid 4 on the PlayStation 3. FJ wants to purchase the set of game consoles (no more than one each) and games (no more than one each -- and within the constraints of a given budget) that helps his cows produce the most milk and thus nourish the most children.
FJ researched N (1 <= N <= 50) consoles, each with a console price Pi (1 <= Pi <= 1000) and a number of console-specific games Gi (1 <= Gi <= 10). A cow must, of course, own a console before she can buy any game that is specific to that console. Each individual game has a game price GPj (1 <= GPj price <= 100) and a production value (1 <= PVj <= 1,000,000), which indicates how much milk a cow will produce after playing the game. Lastly, Farmer John has a budget V (1 <= V <= 100,000) which is the maximum amount of money he can spend. Help him maximize the sum of the production values of the games he buys.

Consider one dataset with N=3 consoles and a V=$800 budget. The first console costs $300 and has 2 games with cost $30 and $25 and production values as shown:
    Game #    Cost    Production Value
      1       $30          50
      2       $25          80

The second console costs $600 and has only 1 game:
    Game #    Cost    Production Value
      1       $50          130

The third console costs $400 and has 3 games:
    Game #    Cost    Production Value
      1       $40         70
      2       $30         40
      3       $35         60

Farmer John should buy consoles 1 and 3, game 2 for console 1, and games 1 and 3 for console 3 to maximize his expected production at 210:
                                Production Value
        Budget:     $800      
        Console 1  -$300
           Game 2   -$25              80
        Console 3  -$400
           Game 1   -$40              70
           Game 3   -$35              60
      -------------------------------------------
        Total:         0 (>= 0)      210

输入描述

• Line 1: Two space-separated integers: N and V
• Lines 2..N+1: Line i+1 describes the price of and the games ?available for console i; it contains: Pi, Gi, and Gi pairs of space-separated integers GPj, PVj

输出描述

• Line 1: The maximum production value that Farmer John can get with his budget.

示例1

输入

3 800 
300 2 30 50 25 80 
600 1 50 130 
400 3 40 70 30 40 35 60 

输出

210

题解

知识点：背包dp。

分组背包套01背包，因为每组物品不能枚举否则超时。

要先开一个临时空间给分组背包内的01背包，因为组内物品是一同考虑的，不能影响一个组的值，最后和原数组没有买游戏机的比较取最大值。

剩下的看代码和01背包差不多。

时间复杂度 \(O(nvg)\)

空间复杂度 \(O(v)\)

代码

#include <bits/stdc++.h>

using namespace std;

int dp[100007], tmp[100007];

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, v;
    cin >> n >> v;
    for (int i = 1, p, g;i <= n;i++) {///组
        cin >> p >> g;
        for (int k = p;k <= v;k++) tmp[k] = dp[k - p];///买游戏机
        for (int j = 1, gp, pv;j <= g;j++) { ///游戏
            cin >> gp >> pv;
            for (int k = v;k >= p + gp;k--)///因为买了游戏机，价格不能小于游戏机加这个游戏
                tmp[k] = max(tmp[k], tmp[k - gp] + pv);///在第i组滚动
        }
        for (int k = p;k <= v;k++) dp[k] = max(dp[k], tmp[k]);///和不买游戏机比
    }
    cout << dp[v] << '\n';
    return 0;
}

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