Partitioning by Palindromes
We say a sequence of characters is a palindrome if it is the same written forwards and backwards. For example, ‘racecar’ is a palindrome, but ‘fastcar’ 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 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. Output For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes. Sample Input 3 racecar fastcar aaadbccb Sample Output 1 7 3
https://odzkskevi.qnssl.com/5cc02c8be522c16812da27d55b5790ce?v=1508140214
【题解】
预处理ok[i][j]表示i到j是否组成回文串，枚举中间点即可
注意偶数回文串的情况
dp[i]表示前i个数的最小划分，转移即可
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cmath> 7 #define max(a, b) ((a) > (b) ? (a) : (b)) 8 #define min(a, b) ((a) < (b) ? (a) : (b)) 9 inline void swap(int &a, int &b) 10 { 11 int tmp = a;a = b;b = tmp; 12 } 13 inline void read(int &x) 14 { 15 x = 0;char ch = getchar(), c = ch; 16 while(ch < '0' || ch > '9')c = ch, ch = getchar(); 17 while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar(); 18 if(c == '-')x = -x; 19 } 20 21 const int INF = 0x3f3f3f3f; 22 const int MAXN = 1000 + 10; 23 24 int t,n,dp[MAXN], ok[MAXN][MAXN]; 25 char s[MAXN]; 26 27 int main() 28 { 29 read(t); 30 for(;t;--t) 31 { 32 memset(ok, 0, sizeof(ok)); 33 scanf("%s", s + 1); 34 n = strlen(s + 1); 35 for(register int i = 1;i <= n;++ i) 36 { 37 int j = i; 38 int tmp = i; 39 while(s[tmp] == s[j] && tmp >= 1 && j <= n) 40 { 41 ok[tmp][j] = 1; 42 -- tmp, ++ j; 43 } 44 tmp = i; 45 j = tmp + 1; 46 while(s[tmp] == s[j] && tmp >= 1 && j <= n) 47 { 48 ok[tmp][j] = 1; 49 -- tmp, ++ j; 50 } 51 } 52 for(register int i = 1;i <= n;++ i) 53 { 54 dp[i] = INF; 55 for(register int j = 1;j <= i;++ j) 56 if(ok[j][i]) dp[i] = min(dp[i], dp[j - 1] + 1); 57 } 58 printf("%d ", dp[n]); 59 } 60 return 0; 61 }