Splits the string

時間限制：1000?ms ?|? 內存限制：65535?KB 描述

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' is a palindrome, but 'abcd' is not.
A partition of a sequence of characters is a list of one or more disjoint non-empty groups of consecutive characters whose concatenation yields the initial sequence. For example, ('race', 'car') is a partition of 'racecar' into two groups.
Given a sequence of characters, we can always create a partition of these characters such that each group in the partition is a palindrome! Given this observation it is natural to ask: what is the minimum number of groups needed for a given string such that every group is a palindrome?
For example:
'racecar' is already a palindrome, therefore it can be partitioned into one group. 'fastcar' does not contain any non-trivial palindromes, so it must be 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 1000 lowercase letters, with no whitespace within.
輸入

Each test case consists of a single line of between 1 and 1000 lowercase letters, with no whitespace within.

輸出

For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.

樣例輸入

racecar fastcar aaadbccb

樣例輸出

1 7 3

題意：給定一個字符串，求最少需要幾個回文子串組成此字符串。

解題思路：用dp[i] 記錄從0到當前i位置這一段最少由幾個回文子串組成，動態轉移方程：dp[i] = min(dp[i],dp[j-1]+1);

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; char str[1010]; int dp[1010]; bool judge(int x,int y) //判斷是不是回文串 {while(x <= y){if(str[x] != str[y])return false;x++;y--;}return true; } int main() {int len, i, j;while(gets(str) != NULL){len = strlen(str);for(i = 0; i < len; i++){dp[i] = i + 1; //假設前面的都不能組成回文串for(j = 0; j <= i; j++)if(str[j] == str[i] && judge(j,i))dp[i] = min(dp[i], dp[j-1]+1);}printf("%d\n",dp[len-1]);}return 0; }

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