ZOJ 3790 Consecutive Blocks 模拟题

Consecutive Blocks

先离散一下，然后模拟，把一种颜色i所在的位置都放入G[i]中，然后枚举一下终点位置，滑动窗体使得起点和终点间花费不超过K，求中间过程的最大值就可以。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define N 100005
set<int>myset;
map<int,int>mp;
set<int>::iterator p;
int n ,k;
int a[N];
vector<int>s[N];
int go(int cur){
	int now = k;
	int ans = 1, l = 0, siz = s[cur].size();
	int len = 1;
	for(int i = 1; i < siz; i++){
		if(s[cur][i-1] == s[cur][i]-1)
		{
			len++;
			ans = max(ans, len);
			continue;
		}
		now -= s[cur][i]-s[cur][i-1]-1;
		if(now>=0)
		{ len++; ans = max(ans, len);}
		else
		{
			while(now<0)
			{
				l++;
				now += s[cur][l]-s[cur][l-1]-1;
				len--;
			}
			len++;
		}
	}
	return ans;
}
int main()
{
	int T ,m,u,v,w, i ,j;
	while(~scanf("%d %d",&n,&k)){
		myset.clear();
		mp.clear();
		for(i=1;i<=n;i++)scanf("%d",&a[i]),myset.insert(a[i]);
		for(p=myset.begin(), i = 1; p!=myset.end(); p++, i++)
			mp.insert(pair<int,int>(*p,i)),s[i].clear();
		int top = i;
		for(i=1;i<=n;i++)
		{
			a[i] = mp.find(a[i])->second;
			s[a[i]].push_back(i);
		}
		int ans = 0;
		for(i=1;i<top;i++)
			ans = max(ans, go(i));
		printf("%d
",ans);		
	}
	return 0;
}
/*
13 3 
1000000000 2 2 1 1 2 2 5 6 8 2 2 2


*/