首页 > 代码库 > UVa 11584 Partitioning by Palindromes (简单DP)
UVa 11584 Partitioning by Palindromes (简单DP)
题意:给定一个字符串,求出它最少可分成几个回文串。
析:dp[i] 表示前 i 个字符最少可分成几个回文串,dp[i] = min{ 1 + dp[j-1] | j-i是回文}。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <unordered_map> #include <unordered_set> #define debug() puts("++++"); #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 5; const int mod = 2000; const int dr[] = {-1, 1, 0, 0}; const int dc[] = {0, 0, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } char s[maxn]; int dp[maxn]; bool judge(int i, int j){ while(i < j){ if(s[i] != s[j]) return false; ++i, --j; } return true; } int main(){ int T; cin >> T; while(T--){ scanf("%s", s); n = strlen(s); memset(dp, INF, sizeof dp); dp[0] = 1; for(int i = 1; i < n; ++i){ for(int j = 0; j < i; ++j) if(judge(j, i)) dp[i] = min(dp[i], 1 + dp[j-1]); else dp[i] = min(dp[i], dp[i-1]+1); } printf("%d\n", dp[n-1]); } return 0; }
UVa 11584 Partitioning by Palindromes (简单DP)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。