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  • CF Preparing Olympiad (DFS)

    Preparing Olympiad
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.

    A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.

    Find the number of ways to choose a problemset for the contest.

    Input

    The first line contains four integers nlrx (1 ≤ n ≤ 15, 1 ≤ l ≤ r ≤ 109, 1 ≤ x ≤ 106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.

    The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 106) — the difficulty of each problem.

    Output

    Print the number of ways to choose a suitable problemset for the contest.

    Sample test(s)
    input
    3 5 6 1
    1 2 3
    output
    2
    input
    4 40 50 10
    10 20 30 25
    output
    2
    input
    5 25 35 10
    10 10 20 10 20
    output
    6




    #include <iostream>
    #include <cstdio>
    #include <string>
    #include <queue>
    #include <vector>
    #include <map>
    #include <algorithm>
    #include <cstring>
    #include <cctype>
    #include <cstdlib>
    #include <cmath>
    #include <ctime>
    #include <climits>
    using    namespace    std;
    
    int    N,L,R,X;
    int    ANS,SUM,DIF,NUM,MAX,MIN = 0x7fffffff;
    int    S[20];
    bool    VIS[20];
    
    void    dfs(int);
    int    main(void)
    {
        cin >> N >> L >> R >> X;
        for(int i = 0;i < N;i ++)
            cin >> S[i];
        sort(S,S + N);
        dfs(0);
        cout << ANS << endl;
    
        return    0;
    }
    
    void    dfs(int start)
    {
        for(int i = start;i < N;i ++)
            if(!VIS[i])
            {
                int    back = DIF;
                int    back_min = MIN;
                int    back_max = MAX;
                MIN = MIN < S[i] ? MIN : S[i];
                MAX = MAX > S[i] ? MAX : S[i];
                VIS[i] = true;
                SUM += S[i];
                DIF = MAX - MIN;
                NUM += 1;
                if(SUM >= L && SUM <= R && NUM >= 2 && DIF >= X)
                    ANS ++;
                if(SUM <= R)
                    dfs(i);
                VIS[i] = false;
                SUM -= S[i];
                DIF = back;
                NUM -= 1;
                MIN = back_min;
                MAX = back_max;
                DIF = MAX - MIN;
            }
    }
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  • 原文地址:https://www.cnblogs.com/xz816111/p/4572784.html
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