http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2631
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:
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.
For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.
3 racecar fastcar aaadbccb
1 7 3题目大意:
找出回文子串的最少数量
解题思路:
状态转移方程:dp[i]=min(dp[j]+1,dp[i]),(j+1,i)是一个回文串
#include <stdio.h> #include <string.h> #include <iostream> using namespace std; int n; char a[1004]; int dp[1111]; bool judge(int x,int y) { /*for(int i=x;i<=y;i++) printf("%c",a[i]); printf("\n");*/ for(int i=x;i<=(x+y)/2;i++) { //printf("(%c %c)\n",a[x],a[y-(i-x)]); if(a[i]!=a[y-(i-x)]) return false; } return true; } int main() { while(~scanf("%d%*c",&n)) { for(int k=0;k<n;k++) { scanf("%s",a+1); int m=strlen(a+1); memset(dp,0x3f3f3f,sizeof(dp)); dp[0]=0; dp[1]=1; for(int i=1;i<=m;i++) for(int j=0;j<i;j++) if(judge(j+1,i)) dp[i]=min(dp[j]+1,dp[i]); printf("%d\n",dp[m]); } } return 0; } /** int main() { int n,m; scanf("%s",a+1); while(scanf("%d%d",&n,&m)) { printf("%d\n",judge(n,m)); } return 0; } **/