【USACO11NOV】牛的阵容Cow Lineup 尺取法+哈希

题目描述

Farmer John has hired a professional photographer to take a picture of some of his cows. Since FJ's cows represent a variety of different breeds, he would like the photo to contain at least one cow from each distinct breed present in his herd.

FJ's N cows are all standing at various positions along a line, each described by an integer position (i.e., its x coordinate) as well as an integer breed ID. FJ plans to take a photograph of a contiguous range of cows along the line. The cost of this photograph is equal its size -- that is, the difference between the maximum and minimum x coordinates of the cows in the range of the photograph.

Please help FJ by computing the minimum cost of a photograph in which there is at least one cow of each distinct breed appearing in FJ's herd.

依次给出N头牛的位置及种类，要求找出连续一段，使其中包含所有种类的牛，问：这连续的一段最小长度是多少？

输入

* Line 1: The number of cows, N (1 <= N <= 50,000).

* Lines 2..1+N: Each line contains two space-separated positive  integers specifying the x coordinate and breed ID of a single  cow.  Both numbers are at most 1 billion.

输出

* Line 1: The smallest cost of a photograph containing each distinct  breed ID.

样例输入

6
25 7
26 1
15 1
22 3
20 1
30 1

样例输出

4

提示

There are 6 cows, at positions 25,26,15,22,20,30, with respective breed IDs 7,1,1,3,1,1.

The range from x=22 up through x=26 (of total size 4) contains each of the distinct breed IDs 1, 3, and 7 represented in FJ's herd.

题解：

迟取法：

设k为总牛数

1.L-R之间牛的等于k,L++.

2.小于k,R++.

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=,MAXN=,M=;
 int gi(){
     int str=;char ch=getchar();
     while(ch>'' || ch<'')ch=getchar();
     while(ch>='' && ch<='')str=str*+ch-'',ch=getchar();
     return str;
 }
 struct Ques{
     int x,id;
 }a[N];
 bool comp(const Ques &p,const Ques &q){
     return p.x<q.x;
 }
 int head[M],num=,ans=MAXN,sum=;
 struct Lin{
     int next,x,cnt;
 }b[N];
 void mark(int x,int t)
 {
     int p=x%M;
     for(int i=head[p];i;i=b[i].next)if(x==b[i].x){
         if(t==)
         {
             if(b[i].cnt==)sum++;
             b[i].cnt++;
         }
         else
         {
             if(b[i].cnt==)sum--;
             b[i].cnt--;
         }
     }
 }
 int Ask(int x)
 {
     int p=x%M;
     for(int i=head[p];i;i=b[i].next)if(x==b[i].x)return b[i].cnt;
     return ;
 }
 void Clear()
 {
     for(int i=;i<=num;i++)b[i].cnt=;
     sum=;
 }
 void init(int x)
 {
     int y=x%M;
     b[++num].next=head[y];
     b[num].x=x;
     b[num].cnt=;
     head[y]=num;
 }
 int main()
 {
     int n=gi(),k=;
     for(int i=;i<=n;i++){
         a[i].x=gi(),a[i].id=gi();
         if(!Ask(a[i].id))k++,init(a[i].id);
     }
     Clear();
     sort(a+,a+n+,comp);
     int l=,r=;
     mark(a[].id,);
     while(l<=r && r<=n)
     {
         if(sum==k){
             ans=min(a[r].x-a[l].x,ans);
             mark(a[l].id,-);
             l++;
         }
         else
         {
             r++;
             mark(a[r].id,);
         }
     }
     printf("%d",ans);
     return ;
 }