題意：輸入一個由小寫字母組成的字符串，你的任務(wù)是把它劃分成盡量少的回文串，字符串長度不超過1000。

分析：

1、dp[i]為字符0~i劃分成的最小回文串的個數(shù)。

2、dp[j] = Min(dp[j], dp[i - 1] + 1)，若i~j是回文串，則更新dp[j]。

#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) const double eps = 1e-8; inline int dcmp(double a, double b){if(fabs(a - b) < eps) return 0;return a > b ? 1 : -1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 1000 + 10; const int MAXT = 10000 + 10; using namespace std; char s[MAXN]; int dp[MAXN]; bool judge(int l, int r){int len = r - l + 1;for(int i = 0; i < len / 2; ++i)if(s[l + i] != s[r - i]) return false;return true; } int main(){int T;scanf("%d", &T);while(T--){memset(dp, INT_INF, sizeof dp);scanf("%s", s + 1);int len = strlen(s + 1);dp[0] = 0;for(int i = 1; i <= len; ++i){for(int j = i; j <= len; ++j){if(judge(i, j)){dp[j] = Min(dp[j], dp[i - 1] + 1);}}}printf("%d\n", dp[len]);}return 0; }

　　

轉(zhuǎn)載于:https://www.cnblogs.com/tyty-Somnuspoppy/p/6417523.html

總結(jié)

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