Блог пользователя Prashant94157

Treasure Hunt problem asked in Trilogy Innovations
Автор Prashant94157, история, 3 года назад, По-английски

In front of you ,there is a treasure containing many gems, soon you realize that they contain bad luck and are not precious as well. You have the power to destroy gems but can destroy 1 gem per sec. You can destroy gems in any order and wanted to destroy maximum bad luck. You have given a 2D matrix, where for i'th gem , it takes arr[i][0] seconds to give arr[i][1] units of bud luck and you can decide in which order you want to receive the bad luck. There is no time lapse in the process. Although while receiving bad luck, you can destroy bad luck. Once gem started giving bad luck , cannot be stopped until it is destroyed of its own after giving bad luck. You have to print maximum bad luck you can destroy.

Constraints

1 <= n <= 10^3 1 <= arr[i][0] <= 10^3 1 <= arr[i][0] <= 10^6

Testcase 1:
Input:
4
0 10
1 4
1 3
1 20
Output: 30
Explanation:
You can get maximum 30 when you choose to destroy 0th and 3rd while receiving bad luck from 1st and 2nd gems. 2 gems require 2 seconds to destroy while 1st and 2nd take 2seconds to give bad luck.

Testcase 2:
Input:
7
3 5
6 5
0 8
8 8
7 4
10 3
8 6
Output:
34
Теги dp
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3 года назад, # |
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It's just a knapsack problem. Please think before asking.

Marinush

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13 месяцев назад, # |
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#include <bits/stdc++.h>
using namespace std;

#define int long long
#define pi pair<int, int>

int mx;

int dp[1001][1001];

int f(vector<vector<int>> &arr,int idx,int rem)
{
    if(rem <= 0) return 0;
    if(idx >= arr.size()) return INT_MAX-mx;
    // take
    if(dp[idx][rem] != -1) return dp[idx][rem];
    int ans1 = arr[idx][1] + f(arr,idx+1,rem-1-arr[idx][0]);
    int ans2 = f(arr,idx+1,rem);

    return dp[idx][rem] = min(ans1,ans2);
}

void solve()
{
    memset(dp,-1,sizeof(dp));
    int n;
    cin>>n;
    mx = 0;
    int total = 0;
    vector<vector<int>> arr(n,vector<int>(2));
    for(int i=0;i<n;i++)
    {
        cin>>arr[i][0]>>arr[i][1];
        mx = max(mx,arr[i][1]);
        total += arr[i][1];
    }
    cout<<total-f(arr,0,n)<<"\n";
}

signed main()
{
    int t;
    t = 1;
    while (t--)
    {
        solve();
    }
}
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    13 месяцев назад, # ^ |
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    // for the Input Input: 7 3 5 6 5 0 8 8 8 7 4 10 3 8 6

    I am getting output : 36 If we can select 10 3 we will get 10 seconds, so we can destroy all other gems ( remaining 6, so required 6 sec < 10 sec). So the answer will be total — 3 => 39-3 = 36

    Right??

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