标签：
Time Limit: 20 Sec
Memory Limit: 256 MB
http://codeforces.com/problemset/problem/6/E
Input
The first line of the input data contains two integer numbers separated by a space n (1 ≤ n ≤ 105) and k (0 ≤ k ≤ 106) — the amount of books by Berlbury in the library, and the maximum allowed height difference between the lowest and the highest books. The second line contains n integer numbers separated by a space. Each number hi (1 ≤ hi ≤ 106) is the height of the i-th book in millimeters.
Output
In the first line of the output data print two numbers a and b (separate them by a space), where a is the maximum amount of books the organizers can include into the exposition, and b — the amount of the time periods, during which Berlbury published a books, and the height difference between the lowest and the highest among these books is not more than k milllimeters.
In each of the following b lines print two integer numbers separated by a space — indexes of the first and the last volumes from each of the required time periods of Berlbury‘s creative work.
Sample Input
3 3
14 12 10
Sample Output
2 2
1 2
2 3
题意
给你一堆数，让你找到最长的区间，使得这个区间里面的最大值减去最小值不超过K，然后让你输出每一个区间的起始位置和结束为止
题解：
类似于双端队列的写法，我们用multiset来处理这个问题，如果当前区间不合法，那么我们不断删去起始端的数就好了
然后不断跑，O(n)
代码
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 1000010 #define mod 10007 #define eps 1e-9 int Num; char CH[20]; //const int inf=0x7fffffff; const int inf=0x3f3f3f3f; /* inline void P(int x) { Num=0;if(!x){putchar(‘0‘);puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } */ inline ll read() { int x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } inline void P(int x) { Num=0;if(!x){putchar(‘0‘);puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } //************************************************************************************** int a[maxn]; vector< pair<int,int> > q; multiset<int> s; int main() { int n=read(),k=read(); for(int i=0;i<n;i++) a[i]=read(); int j=0; int aa=-1; for(int i=0;i<n;i++) { s.insert(a[i]); while(*s.rbegin()-*s.begin()>k) s.erase(s.find(a[j++])); if(i-j+1>aa) { aa=i-j+1; q.clear(); } if(i-j+1==aa) q.push_back(make_pair(j+1,i+1)); } printf("%d %d\n",aa,q.size()); for(int i=0;i<q.size();i++) printf("%d %d\n",q[i].first,q[i].second); }
Codeforces Beta Round #6 (Div. 2 Only) E. Exposition multiset
标签：
原文地址：http://www.cnblogs.com/qscqesze/p/4595525.html
