1.2 milk2 简单模拟

线段，区间。我用的是离散化，在模拟的方法。

线段树应该更高效，有空细看一下线段树

 

Three farmers rise at 5 am each morning and head for the barnto milk three cows. The first farmer begins milking his cow at time300 (measured in seconds after 5 am) and ends at time 1000. Thesecond farmer begins at time 700 and ends at time 1200. The thirdfarmer begins at time 1500 and ends at time 2100. The longestcontinuous time during which at least one farmer was milking a cowwas 900 seconds (from 300 to 1200). The longest time no milking wasdone, between the beginning and the ending of all milking, was 300seconds (1500 minus 1200).

Your job is to write a program that will examine a list ofbeginning and ending times for N (1 <= N <= 5000) farmersmilking N cows and compute (in seconds):

• The longest time interval at least one cow was milked.
• The longest time interval (after milking starts) during whichno cows were being milked.

PROGRAM NAME: milk2

INPUT FORMAT

Line 1: The single integer Lines 2..N+1: Two non-negative integers less than1000000, the starting and ending time in seconds after 0500

SAMPLE INPUT (file milk2.in)

3

300 1000

700 1200

1500 2100/*ID: xdzhbin1LANG: C++TASK:milk2*/#include <fstream>#include <vector>#include <algorithm>using namespace std;//线段开始和结束typedef struct Node{int start;int end;}Node;//自己定义比较函数，sort()函数用bool cmp(Node s, Node e){return s.start < e.start;}int main(){ifstream in("milk2.in",ios_base::in);ofstream out("milk2.out",ios_base::out);int n,i=0;vector<Node> v;in>>n;//读入线段个数nv.resize(n);//为向量申请内存空间，因为现在向量v的size是0for(i=0; i<n; i++){//逐个读入线段开始和结束in>>v[i].start>>v[i].end;}sort(v.begin(),v.end(),cmp);//根据线段开始的地方排序int stop = 0;//最大空闲间隔int from = v[0].start, to = v[0].end; //当前区间开始和结束int cont = to - from;//最大连续区间//逐个扫描线段for(i=1; i<v.size(); i++){if(to > v[i].end)//问题要考虑全面，还有当前扫描线段完全落入当前区间的情况continue;//当前扫描的线段与当前区间相交，更新当前区间if(to >= v[i].start){to = v[i].end;if(to - from > cont)//细心啊，此处调程序是出错，原因是直接更新了cont，没有与之前的比较{cont = to - from;}}else //不相交，也就是当前扫描的线段在当前区间的后面{if(v[i].start - to > stop)//先更新stop信息{stop = v[i].start - to;}from = v[i].start;//把这个线段作为当前区间to = v[i].end;if(to - from > cont)//与之前最大连续区间比较{cont = to - from;}}}out<<cont<<" "<<stop<<endl;in.close();out.close();return 0;}
 

OUTPUT FORMAT

A single line with two integers that represent the longestcontinuous time of milking and the longest idle time.

SAMPLE OUTPUT (file milk2.out)

900 300