zoukankan      html  css  js  c++  java
  • Scientific Conference

    Scientific Conference

    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    Description

    Functioning of a scientific conference is usually divided into several simultaneous sections. For example, there may be a section on parallel computing, a section on visualization, a section on data compression, and so on.
    Obviously, simultaneous work of several sections is necessary in order to reduce the time for scientific program of the conference and to have more time for the banquet, tea-drinking, and informal discussions. However, it is possible that interesting reports are given simultaneously at different sections.
    A participant has written out the time-table of all the reports which are interesting for him. He asks you to determine the maximal number of reports he will be able to attend.

    Input

    The first line contains the number 1 ≤ N ≤ 100000 of interesting reports. Each of the next N lines contains two integers Ts and Teseparated with a space (1 ≤ Ts < Te ≤ 30000). These numbers are the times a corresponding report starts and ends. Time is measured in minutes from the beginning of the conference.

    Output

    You should output the maximal number of reports which the participant can attend. The participant can attend no two reports simultaneously and any two reports he attends must be separated by at least one minute. For example, if a report ends at 15, the next report which can be attended must begin at 16 or later.

    Sample Input

    inputoutput
    5
    3 4
    1 5
    6 7
    4 5
    1 3
    
    3

    贪心:

        结束的越早越是优;

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    const int maxn = 1e5+5;
    using namespace std;
    struct node{
        int s, t;
        bool operator < (const node &x)const{
            return t < x.t;
        }
    }a[maxn];
    int main(){
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++){
            scanf("%d%d", &a[i].s, &a[i].t);
        }
        sort(a, a+n);
        int ans = 0;
        int tmp = 0;
        for(int i = 0; i < n; i++){
            if(a[i].s > tmp){
                tmp = a[i].t;
                ans++;
            }
        }
        printf("%d
    ", ans);
        return 0;
    }
     
  • 相关阅读:
    京东分页
    相册分类列表页
    在线运行Javascript,Jquery,HTML,CSS代码
    点击事件后动画提示
    一些广告代码
    爱可有—之最经典
    爱可有网络社区需要定义
    鼠标移动时缩小图片显示说明
    Flask-RESTful 快速入门
    Sequelize 关系模型简介
  • 原文地址:https://www.cnblogs.com/ACMessi/p/4900651.html
Copyright © 2011-2022 走看看