此页面通过工具从 csdn 导出，格式可能有问题。
题目
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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):
PROGRAM NAME: milk2INPUT FORMAT
SAMPLE INPUT (file milk2.in)3 300 1000 700 1200 1500 2100 OUTPUT FORMATA single line with two integers that represent the longest continuous time of milking and the longest idle time.SAMPLE OUTPUT (file milk2.out)900 300 |
思路
典型的离散化问题，其实也也说不清楚到底什么算是离散化。一般都是长度或者面积上给个区间让你去处理，把复杂的数字映射成自然数，通过 扫描（废话）、合并、裁剪 这么几种来处理“图形”。
所以，这个题，自然而然想到的就是合并线段，把这些线段合起来，看看最长能合出多长，合好之后再看看两两之间的空隙找个最长的然后print出来就OK。
如何合并？一般都是把一段对齐，然后按照另一端的某种顺序排列起来，然后一个一个拾取处理。（突然觉得
Kruskal的本质不也是去合并线段么，只不过合的不是一条线上而是一张图上的，以后研究研究~
）。
对于这个题目，我的处理法子是，把所有的线段按端点位置排序，左端点优先，然后考虑右端点。一个sort而已嘛~ 然后一个一个拾取处理，记录已合成的大线段的起止坐标（l,r表示）。
对于每条线段，有一下三种情况：
|
————
（已合成的线段）
—— （ case 1.1 ） ———— （ case 1.2 ） ———— （ case 2 ） |
即：
1. a[i].l <= r
1) a[i].r <= r 扔掉不用管了。2) a[i].r > r 续上呗~
2. a[i].l > r
更新answer，另起炉灶。
代码
#include <cstdio>
#include <algorithm>
using namespace std;
int n;
struct link{
int l,r;
} a[111111];
int cmp(link a,link b){
return (a.l==b.l)?(a.r<=b.r):(a.l<b.l);
}
int main(){
freopen("milk2.in","r",stdin);
freopen("milk2.out","w",stdout);
scanf("%d",&n);
for (int i(1);i<=n;i++){
scanf("%d%d",&a[i].l,&a[i].r);
}
sort(a+1,a+n+1,cmp);
int l=a[1].l,r=a[1].r,max=a[1].r-a[1].l,nom=0;
for (int i(2);i<=n;i++){
if (a[i].r<=r) continue;
if (a[i].l<=r) {
r=a[i].r;
} else {
if (r-l>max) max=r-l;
if (a[i].l-r>nom) nom=a[i].l-r;
l=a[i].l;
r=a[i].r;
}
}
printf("%d %d\n",max,nom);
return 0;
}
