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  • Anya and Ghosts CodeForces

    Anya and Ghosts
     

    Anya loves to watch horror movies. In the best traditions of horror, she will be visited by m ghosts tonight. Anya has lots of candles prepared for the visits, each candle can produce light for exactly t seconds. It takes the girl one second to light one candle. More formally, Anya can spend one second to light one candle, then this candle burns for exactly t seconds and then goes out and can no longer be used.

    For each of the m ghosts Anya knows the time at which it comes: the i-th visit will happen wi seconds after midnight, all wi's are distinct. Each visit lasts exactly one second.

    What is the minimum number of candles Anya should use so that during each visit, at least r candles are burning? Anya can start to light a candle at any time that is integer number of seconds from midnight, possibly, at the time before midnight. That means, she can start to light a candle integer number of seconds before midnight or integer number of seconds after a midnight, or in other words in any integer moment of time.

    Input

    The first line contains three integers m, t, r (1 ≤ m, t, r ≤ 300), representing the number of ghosts to visit Anya, the duration of a candle's burning and the minimum number of candles that should burn during each visit.

    The next line contains m space-separated numbers wi (1 ≤ i ≤ m, 1 ≤ wi ≤ 300), the i-th of them repesents at what second after the midnight the i-th ghost will come. All wi's are distinct, they follow in the strictly increasing order.

    Output

    If it is possible to make at least r candles burn during each visit, then print the minimum number of candles that Anya needs to light for that.

    If that is impossible, print  - 1.

    Examples
    Input
    1 8 3
    10
    Output
    3
    Input
    2 10 1
    5 8
    Output
    1
    Input
    1 1 3
    10
    Output
    -1

    题意:晚上来了m个厉害的鬼,每个鬼来待一秒,来的时间是固定的,来的时候必须有r根蜡烛是亮的,否则主人公就挂了。主人公有无数根蜡烛,可以在任意时间点亮,每根持
    续亮t秒,每次点亮需要1秒。问一晚上至少需要点亮几根蜡烛。
    大题思路就是模拟,模拟的同时需要考虑最优,就是贪心思想。这道题思路通了写着就很顺了。(弱wa了11次。。。)
    #include<iostream>
    #include<algorithm>
    #include<cstdio>
    #include<string.h>
    using namespace std;
    int time[310];//蜡烛点亮的时间
    int w[310];
    int main()
    123 {
    int m, t, r; cin >> m >> t >> r; for (int i = 1; i <= m; i++) { scanf("%d", &w[i]); } if (t < r) { cout << "-1" << endl; } else { int ans = r; int start = w[1] - r;//最快灭的蜡烛点燃的时间 for (int i = 1; i <= r; i++) { time[i] = start + i - 1; } for (int i = 2; i <= m; i++) { int num = 1; for (int j = 1; j <= r; j++) { if (time[j] + t < w[i])//蜡烛熄灭 { ans++; time[j] = w[i] - num++; } } } cout << ans << endl; } return 0; }
    
    
    
    
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  • 原文地址:https://www.cnblogs.com/Anony-WhiteLearner/p/6336988.html
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