D. Huge Strings Codeforces Round #438 by Sberbank and Barcelona Bootcamp (Div. 1 + Div. 2 combined)(
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http://codeforces.com/contest/868/problem/D
优化:两个串合并
原有状态+ 第一个串的尾部&第二个串的头部的状态
串变为第一个串的头部&第二个串的尾部
注意:
头尾不能重复
如串A合并串A
这就意味着串的头尾不能有重合,
详见代码
#include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <time.h> #include <string> #include <set> #include <map> #include <list> #include <stack> #include <queue> #include <vector> #include <bitset> #include <ext/rope> #include <algorithm> #include <iostream> using namespace std; #define ll long long #define minv 1e-6 #define inf 1e9 #define pi 3.1415926536 #define E 2.7182818284 const ll mod=1e9+7;//998244353 const int maxn=201; int w=15; int f[maxn][1<<(15+1)]={0},add[15+2]; int len,value; string str,s,pre[maxn],post[maxn]; void work(int index,string a) { int i,j,z; len=a.length(); for (j=1;j<=w;j++) { value=0; z=(1<<(j-1))-1; for (i=0;i<len;i++) { value=(value<<1|(a[i]==‘1‘)); if (i>=j-1) { f[index][value+add[j]]=1; value=value&z; } } } } int main() { int n,q,Q,x,y,i; add[1]=0; for (i=2;i<=w+1;i++) add[i]=(1<<i)-2; scanf("%d",&n); for (i=1;i<=n;i++) { cin>>str; work(i,str); len=str.size(); if (len<=w) pre[i]=str; else { pre[i]=str.substr(0,w); if (len>=w+w) post[i]=str.substr(str.length()-w,w); else post[i]=str.substr(str.length()-len+w,len-w); } } scanf("%d",&q); for (Q=n+1;Q<=n+q;Q++) { scanf("%d%d",&x,&y); for (i=0;i<add[w+1];i++) f[Q][i]=f[x][i] | f[y][i]; if (post[x].empty()) str=pre[x]; else if (post[x].length()<w) str=pre[x].substr(pre[x].length()-(w-post[x].length()),w-post[x].length())+post[x]; else str=post[x]; str+=pre[y]; work(Q,str); str=pre[x]+post[x]+pre[y]+post[y]; pre[Q]=str.substr(0,min(w,(int)str.length())); str.erase(0,min(w,(int)str.length())); post[Q]=str.substr(str.length()-min(w,(int)str.length()),min(w,(int)str.length())); for (i=0;i<add[w+1];i++) if (f[Q][i]==0) break; printf("%d ",(int)(log(2+i+minv)/log(2))-1); } return 0; }
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