Similar Number - HDU 4595

Similar Number

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 260    Accepted Submission(s): 62

Problem Description

We can call two numbers “similar” if their lengths are equal and we can make a number ‘’similar’’ through swapping some digits of itself. For instance, 123 is similar to 321 but 12 is not similar to 120.
A sequence can be called “similar sequence” if this sequence has only one pair of ‘’similar’’ numbers.
Now, Zero writes a series of integers and he needs you to answer his queries online. For each query, he gives you two integers l and r, which represent the two endpoints of a segment. He wants to know how many successive subsequences are “similar sequences”. 

 

Input

There are multiple test cases.
In the first line of every test case there are two integers n (0<n<=100000) and m (0<=m<=100000), meaning that the sequence contains n integers and m queries. 
The next line contains n integers a[i] (1<=i<=n, a[i] <= 10⁹), which represent the integers in the sequence.
After that, there are m lines, each line containing two integers l and r.
Zero thinks that it is too simple for him to solve this problem, so he sets a variable k. In the beginning of each test case, k = 0. For each query, he sets l = l + k and r = r - k, Zero guarantees that 0<l<=r<=n, then make k the answer of this query.
The input ends with EOF.

 

Output

For each query, output an integer, and there is no blank line after each test.

 

Sample Input

5 52 2 2 2 2 1 41 81 41 6-1 5

 

Sample Output

31130

 

题意：定义相似数为两个数的数字组合相同，相似区间为这个区间内有且仅有一对相似数。给定l和r，求共有多少个相似区间。

思路：参考http://blog.csdn.net/acm_cxlove/article/details/8931000,大概就是求出区间内相似对数>=1的，和>=2的，然后做差，剩余的就是你=1的

AC代码如下：

#include<cstdio>#include<cstring>#include<map>#include<string>#include<algorithm>using namespace std;typedef long long ll;map<string,int> match;char s[100010][20];int n,m,point[100010],one[100010],two[100010],sum_one[100010],sum_two[100010];bool cmp(char a,char b){    return a<b;}int main(){    int i,j,k,a,b,len,c,pos;    ll l,r,ans;    while(~scanf("%d%d",&n,&m))    {        for(i=1;i<=n;i++)           scanf("%s",s[i]);        match.clear();        for(i=n;i>=1;i--)        {            len=strlen(s[i]);            sort(s[i],s[i]+len,cmp);            k=match[s[i]];            if(k==0)              point[i]=n+1;            else              point[i]=k;            match[s[i]]=i;        }        a=n+1;b=n+1;        for(i=n;i>=1;i--)        {            c=point[i];            if(c<a)              b=a,a=c;            else if(c<b)              b=c;            one[i]=a;            two[i]=b;        }        for(i=1;i<=n;i++)        {            sum_one[i]=sum_one[i-1]+(one[i]-1);            sum_two[i]=sum_two[i-1]+(two[i]-1);        }        ans=0;        while(m--)        {            scanf("%I64d%I64d",&l,&r);            l+=ans;            r-=ans;            ans=0;            pos=upper_bound(one+1,one+n,r)-one-1;            if(pos>=l)            ans+=r*(pos-l+1)-(sum_one[pos]-sum_one[l-1]);            pos=upper_bound(two+1,two+n,r)-two-1;            if(pos>=l)            ans-=r*(pos-l+1)-(sum_two[pos]-sum_two[l-1]);            printf("%I64d\n",ans);        }    }}