HDU4455 Substrings（DP）

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Substrings

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2918 Accepted Submission(s): 895
Problem Description

XXX has an array of length n. XXX wants to know that, for a given w, what is the sum of the distinct elements’ number in all substrings of length w. For example, the array is { 1 1 2 3 4 4 5 } When w = 3, there are five substrings of length 3. They are (1,1,2),(1,2,3),(2,3,4),(3,4,4),(4,4,5)
The distinct elements’ number of those five substrings are 2,3,3,2,2.
So the sum of the distinct elements’ number should be 2+3+3+2+2 = 12

Input

There are several test cases.
Each test case starts with a positive integer n, the array length. The next line consists of n integers a₁,a₂…a_n, representing the elements of the array.
Then there is a line with an integer Q, the number of queries. At last Q lines follow, each contains one integer w, the substring length of query. The input data ends with n = 0 For all cases, 0<w<=n<=10⁶, 0<=Q<=10⁴, 0<= a₁,a₂…a_n <=10⁶

Output

For each test case, your program should output exactly Q lines, the sum of the distinct number in all substrings of length w for each query.

Sample Input

71 1 2 3 4 4 531230

Sample Output

Source

2012 Asia Hangzhou Regional Contest

这题不容易想到，一看题目，看到这数据范围，看到查询的方式。。。一直在往树状数组或者线段树方面去想。想到了用DP解决就不难了。用DP的思路O(n)复杂度解决。以样例为例说明：1 1 2 3 4 4 5；明显dp[1]=n=7;长度为1的时候有7个区间。从长度为1到长度为2，就是把前6个区间往后增加一个数，把最后一个区间去掉。增加的6个数要看在该区间是否出现过，只要看它上一个相等的元素距离是否大于2.所以dp[2]=dp[1]-1+4;

以此类推就可以得出所以的dp值了。dp[i]=dp[i-1]-A+B;减的A是最后一个长度为i-1的区间的不同数的个数，这个很容易预处理得出来。加的B是第t个数到它上一个数的距离大于i-1的个数.这个B值也容易得出。用s[i]表示离上一个数的距离为i的个数，不断减掉就得到B了。

具体看代码：

#include <iostream>#include <stdio.h>#include <math.h>using namespace std;const int MAXN=1000010;int a[MAXN];//1-n输入的数列int f[MAXN];//f[i]表示a[i]在前面最近出现的位置,f[i]==0表示从左到右第一次出现int s[MAXN];//s[i]表示 t-f[t]==i,1<=t<=n的t的个数,即离上一个相等元素的距离为i的个数long long dp[MAXN];//需要输出的结果int ss[MAXN];//ss[i]表示最后的i个数含有的不同元素的个数int main(){    int n;    int m;    while(scanf("%d",&n)==1 && n)    {        memset(f,0,sizeof(f));        memset(s,0,sizeof(s));        //顺着求s数组        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            s[i-f[a[i]]]++;            f[a[i]]=i;        }        memset(f,0,sizeof(f));//f数组标记在后面是否出现过        ss[1]=1;        f[a[n]]=1;        for(int i=2;i<=n;i++)        {            if(f[a[n-i+1]]==0)            {                f[a[n-i+1]]=1;                ss[i]=ss[i-1]+1;            }            else ss[i]=ss[i-1];        }        dp[1]=n;        int sum=n;        //从dp[i-1]扩展到dp[i]就是去掉最后一个区间的个数，把前面的区间长度增加1，        //加上相应增加的种类数        for(int i=2;i<=n;i++)        {            dp[i]=dp[i-1]-ss[i-1];//减掉最后一个区间的种类数            sum-=s[i-1];            dp[i]+=sum;//加上前面的区间增加一个长度后增加的种类数        }        scanf("%d",&m);        int t;        while(m--)        {            scanf("%d",&t);            printf("%I64d\n",dp[t]);        }    }    return 0;}

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