Splits the string time limit:MS | Memory limit:65535 KB Difficulty:3
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Describe
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HRDV is interested in a string,especially the palindrome string. So he wants some palindrome string.
A sequence of characters is a palindrome if it is the same written forwards and backwards. For example, ' Abeba ' was a palindrome, but ' ABCD ' was not.
A partition of sequence of characters are a list of one or more disjoint non-empty groups of consecutive characters whose Concatenation yields the initial sequence. For example, (' Race ', ' car ') was a partition of ' racecar ' into the groups.
Given a sequence of characters, we can always create a partition of these characters such this each group in the partition is a palindrome! Given This observation it's natural to ask:what are the minimum number of groups needed for a Given string such that ever Y group is a palindrome?
For example:
' Racecar ' is already a palindrome, therefore it can being partitioned into one group. ' Fastcar ' does not contain any non-trivial palindromes, so it must is partitioned as (' F ', ' a ', ' s ', ' t ', ' C ', ' A ', ' R '). ' AAADBCCB ' can be partitioned as (' AAA ', ' d ', ' BCCB '). Input begins with the number N of test cases. Each test case consists of a single line of between 1 and lowercase letters, with no whitespace within.
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Input
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Each test case consists of a single line of between 1 and lowercase letters, with no whitespace within.
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Output
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For each test case, output a line containing the minimum number of groups required to partition the input into groups of P Alindromes.
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Sample input
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Racecarfastcaraaadbccb
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Sample output
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173
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Uploaded by
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tc_ Hu Jendong
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#include <iostream>
#include <cstring>
#include <cstdio>
using namespace Std;
int dp[1005];
Char a[1005];
int judge (int i, int j) {
while (a[i] = = A[j]) {
if (i = = j)
Break
i++;
if (i = = j)
Break
j--;
if (i = = j)
Break
// }
if (i = = J && a[i] = = A[j])
return 1;
Else
return 0;
It's too much trouble judging the palindrome.
int mid = (i + j) >> 1;
int m = i;
for (int i; I <= mid; i++) {
if (a[i]! = a[j-i + M])//Note j-i When you add the starting point of the array fragment.
return 0;
}
return 1;
}
int main () {
while (~SCANF ("%s", A + 1)) {
int len = strlen (A + 1);
for (int i = 1; I <= len; i++) {
Dp[i] = i;
for (int j = 1; J <= I; j + +) {
if (a[i] = = A[j] && judge (J, I)) {//dp[i] represents the minimum number of palindrome numbers from 0 to I
Dp[i] = min (Dp[i], dp[j-1] + 1); J from is starting from 0 scanning to I if found J to I is a palindrome,
Then j to I is a palindrome, you can get this recursive relationship
}
}
}
printf ("%d\n", Dp[len]);
}
return 0;
}
16-Minimum Palindrome array
