zoukankan      html  css  js  c++  java
  • Codeforces Round #288 (Div. 2) C. Anya and Ghosts 模拟 贪心

    C. Anya and Ghosts
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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.

    Sample test(s)
    Input
    1 8 3
    10
    Output
    3
    Input
    2 10 1
    5 8
    Output
    1
    Input
    1 1 3
    10
    Output
    -1
    Note

    Anya can start lighting a candle in the same second with ghost visit. But this candle isn't counted as burning at this visit.

    It takes exactly one second to light up a candle and only after that second this candle is considered burning; it means that if Anya starts lighting candle at moment x, candle is buring from second x + 1 to second x + t inclusively.

    In the first sample test three candles are enough. For example, Anya can start lighting them at the 3-rd, 5-th and 7-th seconds after the midnight.

    In the second sample test one candle is enough. For example, Anya can start lighting it one second before the midnight.

    In the third sample test the answer is  - 1, since during each second at most one candle can burn but Anya needs three candles to light up the room at the moment when the ghost comes.

    模拟一下,然后肯定是在这个鬼之前点灯好,越晚点灯越好

    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 1001
    const int inf=0x7fffffff;   //无限大
    int flag[601];
    int a[maxn];
    int kiss[601];
    int main()
    {
        int m,t,r;
        cin>>m>>t>>r;
        for(int i=0;i<m;i++)
        {
            cin>>a[i];
            flag[a[i]+300]=1;
        }
        if(r>t)
            cout<<"-1"<<endl;
        else
        {
            sort(a,a+m);
            for(int i=0;i<601;i++)
            {
                if(flag[i]==1)
                {
                    int check=0;
                    for(int j=i-1;j>=i-t;j--)
                    {
                        if(kiss[j]==1)
                            check++;
                    }
                    for(int j=i-1;j>=i-(r-check);j--)
                        kiss[j]=1;
                }
            }
            int ans=0;
            for(int i=0;i<601;i++)
            {
                if(kiss[i]==1)
                    ans++;
            }
            cout<<ans<<endl;
        }
    }
  • 相关阅读:
    [一个64位操作系统的设计与实现] 3.1 Func_GetFATEntry疑惑
    【参考】 实现X86_64架构下的BootLoader(二)文件系统
    LBA和CHS转换(转)
    Grafana 重置admin密码
    linux-source: not found ubuntu执行脚本报错
    Hbase学习
    高并发理解
    Linux下安装Artemis
    SpringInAction 第八章 发送异步消息
    SpringInAction 六七章总结
  • 原文地址:https://www.cnblogs.com/qscqesze/p/4256181.html
Copyright © 2011-2022 走看看