Partitioning by Palindromes UVA - 11584
2017-08-11 22:50
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题目链接:点我
题意:
给你一个字符串,问最少能分成几个回文串.
思路:
动态规划,dp[i]表示 1 到 i 最少可以分成几个回文串,则dp[i] = min(dp[i], dp[j] + 1) && judge(j,i) : 表示由 j + 1 到 i 能形成回文串.
代码:
4000
题意:
给你一个字符串,问最少能分成几个回文串.
思路:
动态规划,dp[i]表示 1 到 i 最少可以分成几个回文串,则dp[i] = min(dp[i], dp[j] + 1) && judge(j,i) : 表示由 j + 1 到 i 能形成回文串.
代码:
#include<cstring> #include<cstdio> #include<algorithm> #include<cmath> #include<queue> #include<iostream> using namespace std; const int INF = 1e9 + 10; int dp[1000+10]; char ss[1000+10]; bool judge(int l,int r){ while(l < r){ if(ss[l] != ss[r]) return false; l++; r--; }return true; } int main(){ int t; scanf("%d", &t); while(t--){ scanf(" %s", ss+1); int len = strlen(ss+1); for(int i = 1; i <= len; ++i){ dp[i] = INF; for(int j = 0; j < i;++j) if(judge(j+1,i)) dp[i] = min(dp[i], dp[j] + 1); }printf("%d\n",dp[len]); }return 0; }
4000
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