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主题链接：http://115.28.76.232/problem?
pid=1427
Let us consider the sequence a1, a2,..., an of non-negative integer numbers. Denote as ci,j the number of occurrences of the number i among a1,a2,..., aj. We call the sequence k-nice if for all i1<i2 and for all j the following condition is satisfied: ci1,j ≥ ci2,j ?k.
Given the sequence a1,a2,..., an and the number k, find its longest prefix that is k-nice.
10 1 0 1 1 0 2 2 1 2 2 3 2 0 1 0
8 0
用线段树维护 0到A[i]-1间的最小值。用F[A[i]] 统计频率。推断 0 到 A[i]-1范围内的最小值与F[A[i]]-K的大小就可以。
//#pragma comment(linker, "/STACK:36777216")
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <climits>
#include <cassert>
#include <complex>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
#define LL long long
#define MAXN 200100
struct Node{
int left,right,v;
};Node Tree[MAXN<<2];
void Build(int id,int left,int right){
Tree[id].v=0;
Tree[id].left=left;
Tree[id].right=right;
if(left==right) return;
int mid=(left+right)>>1;
Build(id<<1,left,mid);
Build(id<<1|1,mid+1,right);
}
void Update(int id,int pos,int add){
int left=Tree[id].left,right=Tree[id].right;
if(left==right){
Tree[id].v+=add;
return;
}
int mid=(left+right)>>1;
if(mid>=pos)
Update(id<<1,pos,add);
else
Update(id<<1|1,pos,add);
Tree[id].v=min(Tree[id<<1].v,Tree[id<<1|1].v);
}
int Query(int id,int Qleft,int Qright){
int left=Tree[id].left,right=Tree[id].right;
if(left>=Qleft && right<=Qright)
return Tree[id].v;
int mid=(left+right)>>1;
if(mid>=Qright)
return Query(id<<1,Qleft,Qright);
else if(mid<Qleft)
return Query(id<<1|1,Qleft,Qright);
int r1=Query(id<<1,Qleft,Qright),r2=Query(id<<1|1,Qleft,Qright);
return min(r1,r2);;
}
int A[MAXN];
int F[MAXN];
int main(){
int N,K;
while(~scanf("%d%d",&N,&K)){
for(int i=1;i<=N;i++)
scanf("%d",&A[i]);
Build(1,0,N);
memset(F,0,sizeof(F));
int i;
for(i=1;i<=N;i++){
F[A[i]]++;
Update(1,A[i],1);
if(A[i]==0) continue;
int ans=Query(1,0,A[i]-1);
if(ans<F[A[i]]-K) break;
}
printf("%d\n",i-1);
}
}
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原文地址：http://www.cnblogs.com/yxwkf/p/4719905.html
