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  • Priest John's Busiest Day HDU2491 ACM算法设计

    Priest John's Busiest Day

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

    Total Submission(s): 1104    Accepted Submission(s): 306

    Problem Description
    John is the only priest in his town. October 26th is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples plan to get married on the blessed day. The i-th couple plan to hold their wedding from time Si to time Ti. According to the traditions in the town, there must be a special ceremony on which the couple stand before the priest and accept blessings. Moreover, this ceremony must be longer than half of the wedding time and can’t be interrupted. Could you tell John how to arrange his schedule so that he can hold all special ceremonies of all weddings?
    Please note that:
    John can not hold two ceremonies at the same time. John can only join or leave the weddings at integral time. John can show up at another ceremony immediately after he finishes the previous one.
     
    Input
    The input consists of several test cases and ends with a line containing a zero.
    In each test case, the first line contains a integer N ( 1 ≤ N ≤ 100,000) indicating the total number of the weddings.
    In the next N lines, each line contains two integers Si and Ti. (0 <= Si < Ti <= 2147483647)
     
    Output
    For each test, if John can hold all special ceremonies, print "YES"; otherwise, print “NO”.
     
    Sample Input
    3 1 5 2 4 3 6 2 1 5 4 6 0
     
    Sample Output
    NO YES
     
    Source
     
    Recommend
    gaojie
     

    贪心。

    #include <iostream>
    #include <algorithm>
    //#include <fstream>
    using namespace std;
    
    struct Wed
    {    
        int s,t,arrive,leave,timeSpan;
        int operator < (Wed& w)
        {
            if(arrive == w.arrive)
                return t < w.t;
            else 
                return arrive < w.arrive;
        }
    };
    const int MAXn = 100010;
    Wed wed[MAXn];
    
    int main()
    {
        //ifstream cin("in.txt");
        int testCases;
        while (cin>>testCases,testCases)
        {
            //input
            for(int i = 0;i < testCases;i ++)
            {
                cin>>wed[i].s>>wed[i].t;
                wed[i].timeSpan = (wed[i].t - wed[i].s) / 2 + 1;
                wed[i].arrive = wed[i].t - wed[i].timeSpan;
            }
            sort(wed,wed+testCases);
            //Judge
            bool flag = 0;
            wed[0].leave = wed[0].s + wed[0].timeSpan;
            for(int i = 1;i < testCases;i ++)
            {
                if(wed[i].arrive < wed[i-1].leave)
                {
                    flag = 1;
                    break;
                }            
                wed[i].leave = (wed[i-1].leave > wed[i].s ? wed[i-1].leave : wed[i].s) + wed[i].timeSpan;
            }
            //output
            if(flag)
                cout<<"NO"<<endl;
            else
                cout<<"YES"<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ifinver/p/3032591.html
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