p1470 Longest Prefix

原本就想到dp，可是是我的思路是在串的各个位置都遍历一次set，看dp[i-st[k]]是否为1且前length(st[k])是st[k]。这样200000*200*10会超时。更好的办法是在i位取前len<=10个看dp[]和set中是否存在。只要200000*55*log200。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <cassert>
#include <stack>
#include <bitset>
#define mkp make_pair
using namespace std;
const double EPS=1e-8;
typedef long long lon;
const lon SZ=200030,INF=0x7FFFFFFF;
set<string> st;
string str;
int arr[SZ];void init()
{for(;;){string tmp;cin>>tmp;if(tmp==".")break;st.insert(tmp);}string tmp;for(;cin>>tmp;){str+=tmp;}
}void work()
{int res=0;for(int i=0;i<str.size();++i){for(int j=0;j<=9;++j){if(i-j-1<0||arr[i-j-1]){string tmp=str.substr(max(0,i-j),min(i+1,j+1));//if(i==6)cout<<j<<" "<<tmp<<" "<<(st.find(tmp)!=st.end())<<endl;if(st.find(tmp)!=st.end()){arr[i]=1;//cout<<i<<" "<<"here"<<endl;break;}}}if(arr[i]==1){res=i+1;}}cout<<res<<endl;
}int main()
{std::ios::sync_with_stdio(0);//freopen("d:\\1.txt","r",stdin);
    lon casenum;//cin>>casenum;//for(lon time=1;time<=casenum;++time)
    {init();work();}return 0;
}