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Difference Is Beautiful
Description Mr. Flower's business is growing much faster than originally planned. He has now become the CEO of a world-famous beef corporation. However, the boss never lives a casual life because he should take charge of the subsidiary scattered all over the world. Mr. Flower has N companies, and he numbered them with 0 to N – 1. All of the companies will give Mr. Flower a report about the development each year. Among all of the tedious data, only one thing draws Mr. Flower's attention – the turnover. In fact, Mr. Flower will not be angry with the companies running under deficit. He thinks these companies have a large room for future development. What dissatisfy him are those companies who created the same turnover. Because in his eyes, keeping more than Now we know the annual turnover of all companies (an integer sequence Pi, the ith represents the turnover of the ith company this year.). We say a number sequence is perfect if all of its numbers are different from each Input The first line of the input contains two integers N and M. N is the number of companies. M is the number of queries. (1 ≤ N, M ≤ 200000). The second line contains N integer numbers not exceeding Output The output contains M lines. For each query, output the length of the longest consecutive perfect sequence between [L, R] Sample Input 9 2 2 5 4 1 2 3 6 2 4 0 8 2 6 Sample Output 6 5 Hint
The longest perfect sequence of the first query in the sample input is '5 4 1 2 3 6', so the answer for this query is 6.
Source
POJ Monthly--2007.10.06, SHOIT@ZSU
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题意：
给n个数（n<=200000)，每个数的绝对值不超过（10^6），有m个查询（m<=200000),每次查询区间[a,b]中连续的没有相同数的的最大长度。
思路：
先dp预处理，dp[i]表示以i结尾的最长连续不重复数，搞个辅助数组pos[]记录一个点最近一次出现的位置就递推了。
同时也可以处理pre[i]（以i结尾的最长连续不重复串的初始位置）。
RMQ处理dp[]，则对每个询问就可以直接得出答案了，但是会出现边界情况，求出的最长串的初始位置超出了u，故还需要更进一步思考怎样处理边界情况。
对每个询问u、v，二分找到u在pre中的位置pos，那么区间就可以分为两段了[u,pos-1]、[pos,v]，第二段不会出现边界情况，直接RMQ查询，第一段全部都出现边界情况，所以最大值肯定在pos-1这点取了，这样问题就解决了。
思路来源于：男神博客
ps: 这题超经典，强烈推荐.
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
//#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 200005
#define MAXN 1000000
#define mod 1000000000
#define INF 0x3f3f3f3f
using namespace std;
int n,m,ans;
int f[2*maxn][20];
int a[maxn],dp[maxn],pre[maxn];
int pos[2*MAXN+5];
void init_rmq()
{
int i,j;
for(i=1;i<=n;i++)
{
f[i][0]=dp[i];
}
for(j=1;(1<<j)<=n;j++)
{
for(i=1;i+j-1<=n;i++)
{
f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);
}
}
}
int query_rmq(int l,int r)
{
int k=0;
while((1<<(k+1))<=r-l+1) k++;
return max(f[l][k],f[r-(1<<k)+1][k]);
}
void presolve()
{
int i,j;
memset(pos,0,sizeof(pos));
a[0]=-1;
for(i=1;i<=n;i++)
{
if(a[i]==a[i-1]) dp[i]=1;
else dp[i]=min(dp[i-1]+1,i-pos[a[i]]);
pos[a[i]]=i;
pre[i]=i-dp[i]+1;
}
}
void solve()
{
int i,j,t,u,v,pos;
for(i=1;i<=m;i++)
{
scanf("%d%d",&u,&v);
u++,v++;
ans=0;
pos=lower_bound(pre+1,pre+n+1,u)-pre;
if(pos<=v) // 分成两段 [u,pos-1]中的元素各不相同 [pos,v]中的连段段首位置>u 可根据rmq求
{
ans=max(ans,query_rmq(pos,v));
ans=max(ans,pos-u);
}
else ans=max(ans,v-u+1); // pos>v说明u~v中的元素各不相同
printf("%d\n",ans);
}
}
int main()
{
int i,j;
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i]+=MAXN;
}
presolve();
init_rmq();
solve();
}
return 0;
}