洛谷 3029 [USACO11NOV]牛的阵容Cow Lineup

https://www.luogu.org/problem/show?pid=3029

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？

输入输出格式

输入格式：

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.

输出格式：

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

输入输出样例

输入样例#1：

输出样例#1：

说明

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.

感谢 wjcwinmt 提供题目简述

队列

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
int n;
long long ans=2e15;
struct node
{int pos,bl;bool operator < (node p)const{return pos<p.pos;}
}e[50001];
int head,tail,que[50001];
int hassh[50001],sum[50001],cnt;
int main()
{scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d%d",&e[i].pos,&e[i].bl),hassh[i]=e[i].bl;sort(hassh+1,hassh+n+1);int tot=unique(hassh+1,hassh+n+1)-(hassh+1);for(int i=1;i<=n;i++) e[i].bl=lower_bound(hassh+1,hassh+n+1,e[i].bl)-hassh;sort(e+1,e+n+1);for(int i=1;i<=n;i++){if(++sum[e[i].bl]==1) cnt++;que[tail++]=i;while(head<tail && sum[e[que[head]].bl]>1) sum[e[que[head++]].bl]--;if(cnt==tot) ans=min(ans,1ll*e[que[tail-1]].pos-e[que[head]].pos);}cout<<ans;
}

转载于:https://www.cnblogs.com/TheRoadToTheGold/p/7128236.html