Uva--10069（动规，LCS）

2014-08-02 19:31:50

2014-08-02 16:29:05

Problem E

Distinct Subsequences

Input: standard input<style='mso-bidi-font-weight:normal'>

Output: standard output

 

A subsequence of a given sequence is just the given sequence with some elements (possibly none) left out. Formally, given a sequence X=x1x2…xm, another sequence Z=z1z2…zk is a subsequence of X if there exists a strictly increasing sequence <i1, i2, …, ik> of indices of X such that for allj = 1, 2, …, k, we have xij=zj. For example, Z=bcdb is a subsequence of X=abcbdab with corresponding index sequence < 2, 3, 5, 7 >.

In this problem your job is to write a program that counts the number of occurrences of Z in X as a subsequence such that each has a distinct index sequence.

 

Input

The first line of the input contains an integer N indicating the number of test cases to follow.

The first line of each test case contains a string X, composed entirely of lowercase alphabetic characters and having length no greater than 10,000. The second line contains another string Z having length no greater than 100 and also composed of only lowercase alphabetic characters. Be assured that neither Z nor any prefix or suffix of Z will have more than 10100 distinct occurrences in X as a subsequence.

 

Output

For each test case in the input output the number of distinct occurrences of Z in X as a subsequence. Output for each input set must be on a separate line.

 

Sample Input

2
babgbag
bag
rabbbit
rabbit

 

Sample Output

5
3

思路：这道DP + 高精度，坑我到3点 TAT（还让不让人睡了！），不得不说一下，首先看了下题目，结果比较大（10^100级），果断贴了以前敲的大数模板，竟然RE！无奈把数组改成vector才过，QAQ！

DP的方式有两种：（1）以string 1为基准（也就是放在第一层循环里），扫string 2 ，这种方法比较常见，dp[i][j] 表示子串前 j 个组成的字符串在母船前 i 个组成字符串中的出现次数。

　　　　　　　　　　DP方程：(1) s1[i] == s2[j] : dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] ;

　　　　　　　　　　　　　　 (2) s1[i] != s2[j] : dp[i][j] = dp[i - 1][j];

　　　　　　　　（2）以string 2 为基准（也就是放在第一层循环里)，扫string 1，我是用这种方式来实现的，有些奇葩，但是方程是我自己推的。

　　　　　　　　　　DP方程：

　　　　　　　　　　if(s1[i - 1] == s2[j - 1]){
                     if(j - 1 > 0 && s2[j - 1] == s2[j - 2])
                         dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                     else
                         dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
                 }
　　　　　　　　　　else
　　　　　　　　　　　　　　dp[i][j] = dp[i - 1][j];
  1 /*************************************************************************
  2     > File Name: i.cpp
  3     > Author: Nature
  4     > Mail: 564374850@qq.com
  5     > Created Time: Thu 31 Jul 2014 12:34:01 AM CST
  6 ************************************************************************/
  7 
  8 #include <cstdio>
  9 #include <cstring>
 10 #include <cstdlib>
 11 #include <cmath>
 12 #include <vector>
 13 #include <iostream>
 14 #include <algorithm>
 15 using namespace std;
 16 #define ll long long
 17 const int maxn = 1000;
 18 
 19 struct bign{
 20     int len;
 21     vector<int> s;
 22 
 23     //初始化与赋值
 24     bign(){
 25         s.clear();
 26         len = 0;//因为0的存在
 27     }
 28 
 29     bign operator = (const char *num){
 30         len = strlen(num);
 31         for(int i = 0; i < len; ++i)
 32             s.push_back(num[len - i - 1] - '0');
 33             //s[i] = num[len - i - 1] - '0';//倒序转化成数字
 34         return *this;
 35     }
 36 
 37     bign operator = (int num){
 38         char ts[maxn + 5];
 39         sprintf(ts,"%d",num);
 40         *this = ts;
 41         return *this;
 42     }
 43 
 44     bign(int num){
 45         *this = num;
 46     }
 47 
 48     //bign的str型转化
 49     string str() const{
 50         string res = "";
 51         for(int i = len - 1; i >= 0; --i)
 52             res +=  (char)(s[i] + '0');
 53         if(res == "")
 54             res = "0";
 55         return res;
 56     }
 57     //运算符重载
 58     //高精度加
 59     bign operator + (const bign & b) const{
 60         bign sum;
 61         sum.len = 0;
 62         for(int i = 0, g = 0; g || i < max(len,b.len); ++i){
 63             int x = g;//x:暂存和,g:进位
 64             if(i < len) x += s[i];
 65             if(i < b.len) x += b.s[i];
 66             sum.s.push_back(x % 10);
 67             sum.len++;
 68             g = x / 10;
 69         }
 70         return sum;
 71     }
 72 
 73     friend ostream & operator << (ostream & out,const bign & x);
 74 };
 75 
 76 ostream & operator << (ostream & out,const bign & x){
 77     out << x.str();
 78     return out;
 79 }
 80 
 81 bign dp[10005][105];
 82 
 83 int main(){
 84     //freopen("in","r",stdin);
 85     int Case;
 86     int len1,len2;
 87 
 88     char s1[10005];
 89     char s2[105];
 90     scanf("%d",&Case);
 91     getchar();
 92     while(Case--){
 93         gets(s1);
 94         gets(s2);
 95         len1 = strlen(s1);
 96         len2 = strlen(s2);
 97         for(int j = 1; j <= len2; ++j){
 98             for(int i = 1; i <= len1; ++i){
 99                 dp[i][0] = 1;
100                 if(s1[i - 1] == s2[j - 1]){
101                     if(j - 1 > 0 && s2[j - 1] == s2[j - 2])
102                         dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
103                     else
104                         dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
105                 }
106                 else
107                     dp[i][j] = dp[i - 1][j];
108             }
109         }
110         cout << dp[len1][len2] << endl;
111     }
112     return 0;
113 }

 

 
posted @ 2014-08-02 19:32  Naturain  阅读(167)  评论(0)  编辑  收藏  举报