USACO 1.2 Milking Cows

题目连接

http://train.usaco.org/usacoprob2?a=ONvqe5bkAcm&S=milk2

Description

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

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

The longest time interval at least one cow was milked.
The longest time interval (after milking starts) during which no cows were being milked.

Input

Line 1: The single integer, N
Lines 2..N+1: Two non-negative integers less than 1,000,000, respectively the starting and ending time in seconds after 0500

Output

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

Sample Input

3
300 1000
700 1200
1500 2100

Sample Output

900 300

题意

给出每头牛的工作时间段,输出最长至少一头牛的工作时间长度以及最长没有妞工作的时间长度。

题解

时间按开始时间早晚排序,用cur维护当前最长的从头到尾的时间段,若开始时间在cur的结束时间之前,且当前时间段的结束时间>cur的结束时间,则更新cur的结束时间为当前时间的结束时间,否则如果当前时间段开始时间>=cur的结束时间,而且结束时间>=cur的结束时间,说明这个时间段在cur之外,则维护tnomilk的值和tmilk的值,且令这个时间段成为cur。

代码

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/*
ID: hyson601
PROG: milk2
LANG: C++11
*/
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=1e4+10;
int nmilking,tmilk,tnomilk,t;
struct node{
int begin;
int end;
friend bool operator <(node a,node b){
return a.begin<b.begin;
}
}milking[maxn],cur;
int main(){
freopen("milk2.in", "r", stdin);
freopen("milk2.out", "w", stdout);
scanf("%d",&nmilking);
for(int i=0;i<nmilking;++i)
scanf("%d%d",&milking[i].begin,&milking[i].end);
sort(milking,milking+nmilking);
tmilk=0,tnomilk=0;
cur=milking[0];
for(int i=1;i<nmilking;++i){
if(milking[i].begin>cur.end){
t=milking[i].begin-cur.end;
if(t>tnomilk)
tnomilk=t;
t=cur.end-cur.begin;
if(t>tmilk)
tmilk=t;
cur=milking[i];
}else{
if(milking[i].end>cur.end)
cur.end=milking[i].end;
}
}
t=cur.end-cur.begin;
if(t>tmilk)
tmilk=t;
printf("%d %d\n",tmilk,tnomilk);
return 0;
}