UVa 1401 Remember The Word

Trie+dp

Terry Petition training on the problem.

It says that Italy is a string that can have a number of small strings composed.

For example

Abcd

4

A

B

Cd

Ab

ABCD=A+B+CD. ABCD=AB+CD;

Recursion is: From the last one forward. dp[i]=dp[i]+dp[i+ Len[x]] x is appended to the node in the order it was entered. is the prefix for I~strlen (S). s[1,2,3,..., i,... Len]

When building the trie tree, the order is also appended to the node.

At the end of the search, find out how many strings are prefixed and add up the prefixes.

#include <cstdio> #include <cstring> #include <string> #include <queue> #include <algorithm > #include <map> #include <stack> #include <iostream> #include <list> #include <set># Include<cmath> #define INF 0x7fffffff#define EPS 1e-6#define LL long longusing namespace Std;const int mod=20071027;    struct trie{int word[4001*100][26];    int ex[4001*100];    int size;        void Clear () {memset (word[0],0,sizeof (word[0]));        size=1;    ex[0]=0;        } int Insert (char *s,int v) {int u=0,c;            for (int i=0; s[i]!= '; i++) {int c=s[i]-' a ';                if (!word[u][c]) {memset (word[size],0,sizeof (word[size]));                ex[size]=0;            word[u][c]=size++;        } U=word[u][c];    } ex[u]=v;        } int Search (char *s,int len,vector<int>& ans) {int u=0,c;    for (int i=0; s[i]!= ' &&i<len; i++)    {c=s[i]-' a ';            if (!word[u][c]) return 0;            U=WORD[U][C];        if (Ex[u]) ans.push_back (Ex[u]);    }}} Wo;char str[300001];int dp[300001];int n;int le[4001];int main () {int nn=1;        while (scanf ("%s", str)!=eof) {scanf ("%d", &n);        Wo.clear ();        Memset (Dp,0,sizeof (DP));        Char tmp[101];        int Len=strlen (str);            for (int i=0; i<n; i++) {scanf ("%s", TMP);            Le[i+1]=strlen (TMP);        Wo.insert (tmp,i+1);        } dp[len]=1;            for (int i=len-1; i>=0; i--) {vector<int>ans;            Wo.search (Str+i,len-i,ans);        for (int j=0; j<ans.size (); j + +) dp[i]= (Dp[i]+dp[i+le[ans[j]])%mod;    } printf ("Case%d:%d\n", nn++,dp[0]); }}

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UVa 1401 Remember The Word