HDU 4455 Substrings[多重dp]

最新推荐文章于 2017-10-30 20:21:22 发布

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最新推荐文章于 2017-10-30 20:21:22 发布

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文章标签： java

原文链接：http://www.cnblogs.com/shenben/p/6721579.html

本文介绍了一种解决特定子串问题的方法，该问题要求计算给定长度的所有子串中不同元素的数量总和。通过离散化和动态规划技术，文章提供了一种高效的解决方案。

Substrings

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3269 Accepted Submission(s): 999

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 a ₁,a ₂…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<=10 ⁶, 0<=Q<=10 ⁴, 0<= a ₁,a ₂…a _n <=10 ⁶

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

7 1 1 2 3 4 4 5 3 1 2 3 0

Sample Output

7 10 12

Source

2012 Asia Hangzhou Regional Contest

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#include<cstdio>
#include<cstring>
using namespace std;
const int N=1e6+5;
int n,num[N];
int cx[N],add[N],lnc[N];
long long f[N];
void Discretization(){
    memset(cx,0,sizeof cx);
    memset(lnc,0,sizeof lnc);
    for(int i=1;i<=n;i++){
        if(!cx[num[n-i+1]]){
            cx[num[n-i+1]]=1;
            lnc[i]=lnc[i-1]+1;
        }
        else{
            lnc[i]=lnc[i-1];
        }
    }
    memset(cx,0,sizeof cx);
    memset(add,0,sizeof add);
    for(int i=1;i<=n;i++){
        add[i-cx[num[i]]]++;
        cx[num[i]]=i;
    }
}
void DynamicProgramming(){
    memset(f,0,sizeof f);
    f[1]=n;int delta=n;
    for(int i=2;i<=n;i++){
        f[i]=f[i-1]-lnc[i-1];
        delta-=add[i-1];
        f[i]+=delta;
    }
}
void Solution(){
    int Q,x;
    for(scanf("%d",&Q);Q--;) scanf("%d",&x),printf("%lld\n",f[x]);
}
int main(){
    while((~scanf("%d",&n))&&n){
        for(int i=1;i<=n;i++) scanf("%d",&num[i]);
        Discretization();
        DynamicProgramming();
        Solution();
    }
    return 0;
}