usaco 2.3 prefix

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Dynamic programming动态规划

由前往后面扫S字符串

dp[i+prim[j].size()-1 ] = dp[i-1] + prim[j].size() (prim[j]与s.substr(i,i+prim[j].size() 匹配))

加上一个限制条件dp[i-1]!=0,则可以保证可以与前缀连接起来

#include <iostream>#include <cstdio>#include <cstring>#define N 202using namespace std;string s = "";string prim[N];int ps = 0;int dp[200020];int main(){    freopen("prefix.in","r",stdin);    freopen("prefix.out","w",stdout);    int maxsub = 0;    do    {        cin >> prim[ps];        if (prim[ps].size()>maxsub) maxsub = prim[ps].size();    }while (prim[ps++][0]!='.');    string tmp;    while (cin >> tmp)    {        s +=tmp;    }    memset(dp,0,sizeof(dp));    for (int j=0;j<ps;j++)    {        tmp = s.substr(0,prim[j].size());        if (tmp==prim[j])        {            dp[0+prim[j].size()-1] = prim[j].size();        }    }    for (int i=1;i<s.size();i++)    {        for (int j=0;j<ps;j++)        {            //如果剩余匹配长度大于基元长度，则无必要比较            if (prim[j].size()+i>s.size()) continue;            //如果不是前缀子串，则没必要比较            if (dp[i-1]==0) continue;            tmp = s.substr(i,prim[j].size());            if (tmp==prim[j])            {                if (prim[j].size()+dp[i-1] > dp[i+prim[j].size()-1])                dp[i+prim[j].size()-1] = prim[j].size()+dp[i-1];            }        }    }    maxsub = 0;    for (int i=0;i<s.size();i++)    {        if (dp[i]>maxsub) maxsub = dp[i];    }    cout << maxsub << endl;    return 0;}