hdu 6068--Classic Quotation(kmp+DP)

题目链接

 

Problem Description
When online chatting, we can save what somebody said to form his ''Classic Quotation''. Little Q does this, too. What's more? He even changes the original words. Formally, we can assume what somebody said as a string S whose length is n. He will choose a continuous substring of S(or choose nothing), and remove it, then merge the remain parts into a complete one without changing order, marked as S. For example, he might remove ''not'' from the string ''I am not SB.'', so that the new string S will be ''I am SB.'', which makes it funnier.



After doing lots of such things, Little Q finds out that string T occurs as a continuous substring of S very often.

Now given strings S and T, Little Q has k questions. Each question is, given L and R, Little Q will remove a substring so that the remain parts are S[1..i] and S[j..n], what is the expected times that T occurs as a continuous substring of S if he choose every possible pair of (i,j)(1iL,Rjn) equiprobably? Your task is to find the answer E, and report E×L×(nR+1) to him.

Note : When counting occurrences, T can overlap with each other.
 

 

Input
The first line of the input contains an integer C(1C15), denoting the number of test cases.

In each test case, there are 3 integers n,m,k(1n50000,1m100,1k50000) in the first line, denoting the length of S, the length of T and the number of questions.

In the next line, there is a string S consists of n lower-case English letters.

Then in the next line, there is a string T consists of m lower-case English letters.

In the following k lines, there are 2 integers L,R(1L<Rn) in each line, denoting a question.
 

 

Output
For each question, print a single line containing an integer, denoting the answer.
 

 

Sample Input
1
8 5 4
iamnotsb
iamsb
4 7
3 7
3 8
2 7
 

 

Sample Output
1
1
0
0
 
 
题意:有小写字符串s和t,现在在s中去掉连续子串后,剩余s[1…i] 和 s[j…n] 连在一起构成一个新s串,计算t串在新s串中出现了几次。现在q次询问,每次输入L和R,去掉连续子串后s[1…i]和s[j...n]拼接成新串s,1<=i<=L && R<=j<=n,求t串在这些新串中出现的次数和?
 
思路:
          
 
 
代码如下:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int N=50005;
char s[N],t[105];
int pre[N],num[N][105];
int suf[N][105];
int next1[105];
int next2[105][30],flag[105][30];
int n,m,q;

void KMP()
{
    next1[0]=0;
    for(int i=1,k=0; i<m; ++i)
    {
        while(k>0 && t[i]!=t[k]) k=next1[k-1];
        if(t[i]==t[k]) k++;
        next1[i]=k;
    }
}

void cal()
{
    memset(flag,0,sizeof(flag));
    for(int i=0;i<m;i++)
    {
        for(int j=0;j<26;j++)
        {
            char x=j+'a';
            int k=i;
            while(k>0 && t[k]!=x) k=next1[k-1];
            if(t[k]==x) k++;
            next2[i][j]=k;
            if(k==m) flag[i][j]=1,next2[i][j]=next1[m-1];
        }
    }

    memset(pre,0,sizeof(pre));
    memset(num,0,sizeof(num));
    for(int i=0,k=0;i<n;i++)
    {
        while(k>0&&t[k]!=s[i]) k=next1[k-1];
        if(t[k]==s[i]) k++;
        if(k==m) pre[i]++,num[i][next1[m-1]]=1;
        else num[i][k]=1;
        pre[i]+=pre[i-1];
    }
    for(int i=1;i<n;i++)
        for(int j=0;j<m;j++)
            num[i][j]+=num[i-1][j];
    for(int i=1;i<n;i++) pre[i]+=pre[i-1];///前缀和;

    memset(suf,0,sizeof(suf));
    for(int i=n-1;i>=0;i--)
    {
        int x=s[i]-'a';
        for(int j=0;j<m;j++)
            suf[i][j]=flag[j][x]+suf[i+1][next2[j][x]];
    }
    for(int j=0;j<m;j++) ///后缀和;
       for(int i=n-1;i>=0;i--)
          suf[i][j]+=suf[i+1][j];
}

int main()
{
    int T; cin>>T;
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&q);
        scanf("%s%s",s,t);
        KMP();
        cal();
        while(q--)
        {
            int L,R; scanf("%d%d",&L,&R);
            LL ans=(LL)pre[L-1]*(LL)(n-R+1);
            for(int i=0;i<m;i++)
            {
                ans+=(LL)num[L-1][i]*(LL)suf[R-1][i];
            }
            printf("%lld\n",ans);
        }
    }
    return 0;
}
/**
2342
8 3 3463
abcababc
abc
8 3 234
aabbcccbbb
aaabb

4
10 3 23
ababcababc
aba
3 5
*/

 

 
posted @ 2017-09-08 21:24 茶飘香~ 阅读(...) 评论(...) 编辑 收藏