[Atcoder CODE FESTIVAL 2017 qual C]D - Yet Another Palindrome Partitioning 状压DP
2017-10-24 07:58
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题目意思就是划分成N个子串,每个子串中出现奇数次的字符最多为1个,最小化N。
设a[i]=s[1]xor s[2] xor … xor s[i]。那么一个串[l,r]可以被划分出来当a[r]xor a[l-1]=0或2的幂。
N^2的DP方程:F[i]=min{F[j]}+1(a[i]xor a[j-1]==0或2的幂)
但我们发现a[i]只有2^26种,于是设G[k]=min{F[i]} (a[i]==k)
于是从左往右DP,对于每一个G[a[i]]只有27种转移,即min{G[a[i]],G[a[i] xor 2^c]} (0<=c<=25)。最后答案即为G[a
]。时间复杂度O(n*26)。
官方题解更好理解:
代码:
设a[i]=s[1]xor s[2] xor … xor s[i]。那么一个串[l,r]可以被划分出来当a[r]xor a[l-1]=0或2的幂。
N^2的DP方程:F[i]=min{F[j]}+1(a[i]xor a[j-1]==0或2的幂)
但我们发现a[i]只有2^26种,于是设G[k]=min{F[i]} (a[i]==k)
于是从左往右DP,对于每一个G[a[i]]只有27种转移,即min{G[a[i]],G[a[i] xor 2^c]} (0<=c<=25)。最后答案即为G[a
]。时间复杂度O(n*26)。
官方题解更好理解:
代码:
#include<cstdio> #include<iostream> #include<cstring> #define chkmin(a,b) a=min(a,b); using namespace std; const int maxn=300010; int n,f[70000000],h[maxn]; char s[maxn]; int main() { scanf("%s",s+1);n=strlen(s+1); memset(f,0x3f,sizeof(f));f[0]=0; for(int i=1;i<=n;i++) 4000 h[i]=h[i-1]^(1<<(s[i]-'a')); for(int i=1;i<=n;i++) for(int j=0;j<26;j++) chkmin(f[h[i]],f[h[i]^(1<<j)]+1); printf("%d",max(1,f[h ])); }
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