[luogu1090 SCOI2003] 字符串折疊(區間DP+hash)
阿新 • • 發佈:2018-10-04
cst scan clas sca scoi2003 斷點 res const www
傳送門
Solution
區間DP,枚舉斷點,對於一個區間,枚舉折疊長度,用hash暴力判斷是否能折疊即可
Code
#include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #define Re register #define F(i,a,b) for(Re int i=(a);i<=(b);i++) using namespace std; typedef unsigned long long ull; inline int read() { int x=0,f=1;char c=getchar(); while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();} while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar(); return x*f; } const int N=110,bas=137; int n; ull hs[N],pw[N]; char s[N]; int dp[N][N]; ull get_h(int l,int r) {return hs[r]-hs[l-1]*pw[r-l+1];} int get_w(int x) {int res=0;while(x) x/=10,res++;return res;} int main() { scanf("%s",s+1);n=strlen(s+1); F(i,1,n) hs[i]=hs[i-1]*bas+s[i]; pw[0]=1; F(i,1,n) pw[i]=pw[i-1]*bas; memset(dp,0x3f,sizeof(dp)); F(i,1,n) dp[i][i]=1; F(len,2,n) for(Re int i=1,j=len;j<=n;i++,j++) { F(k,i,j-1) dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]); F(k,1,len) if(len%k==0) { ull val=get_h(i,i+k-1);bool fla=0; for(int l=i+k;l<=j;l+=k) if(get_h(l,l+k-1)!=val) {fla=1;break;} if(!fla) dp[i][j]=min(dp[i][j],2+dp[i][i+k-1]+get_w(len/k)); } } printf("%d",dp[1][n]); return 0; }
[luogu1090 SCOI2003] 字符串折疊(區間DP+hash)