HDU 4632 Palindrome subsequence (区间dp 容斥定律)
HDU 4632 Palindrome subsequence (区间dp 容斥定理)
Total Submission(s): 2610 Accepted Submission(s): 1050
Problem Description
Palindrome subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 2610 Accepted Submission(s): 1050
Problem Description
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D>
is a subsequence of <A, B, C, D, E, F>.
(http://en.wikipedia.org/wiki/Subsequence)
Given a string S, your task is to find out how many different subsequence of S is palindrome. Note that for any two subsequence X = <Sx1, Sx2, ..., Sxk> and Y = <Sy1, Sy2, ..., Syk> , if there exist an integer i (1<=i<=k) such that xi != yi, the subsequence X and Y should be consider different even if Sxi = Syi. Also two subsequences with different length should be considered different.
(http://en.wikipedia.org/wiki/Subsequence)
Given a string S, your task is to find out how many different subsequence of S is palindrome. Note that for any two subsequence X = <Sx1, Sx2, ..., Sxk> and Y = <Sy1, Sy2, ..., Syk> , if there exist an integer i (1<=i<=k) such that xi != yi, the subsequence X and Y should be consider different even if Sxi = Syi. Also two subsequences with different length should be considered different.
Input
The first line contains only one integer T (T<=50), which is the number of test cases. Each test case contains a string S, the length of S is not greater than 1000 and only contains lowercase letters.
Output
For each test case, output the case number first, then output the number of different subsequence of the given string, the answer should be module 10007.
Sample Input
4 a aaaaa goodafternooneveryone welcometoooxxourproblems
Sample Output
Case 1: 1 Case 2: 31 Case 3: 421 Case 4: 960
Source
2013 Multi-University Training Contest 4
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4632
题目大意：给一个字符串，求其所有回文子串的个数，字符位置不同属于不同的回文子串
题目分析：很熟悉的一道dp，貌似是编程之美资格赛的一题，原来是曾经多校的原题啊。。。我们从左向右枚举区间，对于每个区间从外向内算，dp[i][j]表示从i到j的子串中回文子串的个数如果s[i] == s[j]，则dp[i][j] += dp[i - 1][j + 1] + 1，另外dp[i][j] = dp[i - 1][j] + dp[i][j + 1] - dp[i - 1][j + 1]，这里相当于一个容斥，去掉重复的部分，最后的答案为dp[len][1]
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4632
题目大意：给一个字符串，求其所有回文子串的个数，字符位置不同属于不同的回文子串
题目分析：很熟悉的一道dp，貌似是编程之美资格赛的一题，原来是曾经多校的原题啊。。。我们从左向右枚举区间，对于每个区间从外向内算，dp[i][j]表示从i到j的子串中回文子串的个数如果s[i] == s[j]，则dp[i][j] += dp[i - 1][j + 1] + 1，另外dp[i][j] = dp[i - 1][j] + dp[i][j + 1] - dp[i - 1][j + 1]，这里相当于一个容斥，去掉重复的部分，最后的答案为dp[len][1]
#include <cstdio>
#include <cstring>
int const MAX = 1005;
int const MOD = 10007;
char s[MAX];
int dp[MAX][MAX];
int main()
{
int T;
scanf("%d", &T);
for(int ca = 1; ca <= T; ca++)
{
printf("Case %d: ", ca);
memset(dp, 0, sizeof(dp));
scanf("%s", s + 1);
int len = strlen(s + 1);
for(int i = 1; i <= len; i++)
{
for(int j = i; j >= 1; j--)
{
if(i == j)
{
dp[i][j] = 1;
continue;
}
if(s[i] == s[j])
dp[i][j] += dp[i - 1][j + 1] + 1;
dp[i][j] += (5 * MOD + dp[i - 1][j] + dp[i][j + 1] - dp[i - 1][j + 1]) % MOD;
}
}
printf("%d\n", dp[len][1] % MOD);
}
}
版权声明：本文为博主原创文章，未经博主允许不得转载。