LA 3942 && UVa 1401 Remember the Word (Trie + DP)

Test instructions: give you a dictionary with a different word of s and a long string l, so that you can break the long string into several words (the word is reusable) and how many. (Algorithm starter Training guide-p209)

Analysis: I go, a look at this is not a DP it? Just started to pay has been runtime error, looking for a long time, always thought is the array opened small, constantly increasing or so, later found that I used the char type ... The following analysis of the topic

It should not be difficult to think of this state transfer equation:

D (i) = Sum{d (I+len (x)) | Word X is s[i ... L] of the prefix}, where Len (x) is the length. D (i) represents a string starting from character I (that is, suffix s[i ...). L]) Number of species.

Obviously we are pushing forward from the back, and the D (i) species should be made up of D (i) and D (I+len (x)) and composed (think why, do not understand can draw a diagram analysis).

If you enumerate x first, then judge if it is s[i ... L] prefix, the time complexity is too high. So change a train of thought, first make the word tire (prefix tree), then try to find s[i in tire. L]. Specific reference code.

The code is as follows:

#include <iostream>#include<cstdio>#include<cstring>using namespacestd;Const intMAXN =4000* -+Ten;//4,000 words, each word the longest is 100, at most there are so manyConst intMAXM =300010;Const intMoD =20071027;intD[MAXM];CharSS[MAXM], t[ the];structtire{intch[maxn][ -]; intVAL[MAXN]; intsz; voidInit () {sz =1; Memset (Val,0,sizeof(Val)); memset (ch[0],0,sizeof(ch[0])); }//Initialize    intIdxCharc) {returnA O'a'; }//Get number    voidInser (Char*s) {//Insert        intU =0, n =strlen (s);  for(inti =0; I < n; ++i) {            intc =idx (s[i]); if(!Ch[u][c]) {memset (Ch[sz],0,sizeof(Ch[sz])); CH[U][C]= sz++; } u=Ch[u][c]; } Val[u]=N; }    voidQuary (Char*s,intIintN) {//Find        intU =0;  for(intj =0; J < N; ++j) {            intc =idx (s[j]); if(!ch[u][c])return ; U=Ch[u][c]; if(Val[u]) d[i] = (D[i] + d[i + val[u])%MoD; }    }}; Tire Tire;intMain () {intN, Kase =0;  while(~SCANF ("%s%d", SS, &N)) {tire.init ();//Be sure to initialize, just start tire definition in the inside, a run on the collapse ... I'm drunk, too.                 for(inti =0; I < n; ++i) {scanf ("%s", T);        Tire.inser (t); } memset (d,0,sizeof(d)); intLen =strlen (ss); D[len]=1;//This place is the boundary, notice the initialization         for(inti = len-1; I >=0; --i) tire.quary (SS+i, I, len-i);//Recursive seed Countprintf"Case %d:%d\n", ++kase, d[0]); }    return 0;}

LA 3942 && UVa 1401 Remember the Word (Trie + DP)