Poj1625 censored! AC automation + dp + High Precision

Problem address: http://poj.org/problem? Id = 1625

 

[Preface]

This is my 100th question on poj.

I did 99 the day before yesterday. I wanted to find one and complete 100, but I didn't find the result.

Yesterday I found an automatic machine + dp.

Return to WA after submission.

The addition operation was rewritten only when it was learned that it was of high precision.

An AC is returned, successfully reaching 100.

Yesterday, csdn said that it was a one-time maintenance service and could not issue a problem-solving report.

I would like to commemorate the 100th th question.

 

[Idea]

In the same way as in the previous article, the AC automatic mechanism is constructed, including the construction of null nodes and failure pointers, and the so-called substrings that are invalid strings are also invalid strings.

Then you can start DP.

The same is true.

The difference is that the transfer equation is DP [I] [J] = DP [I] [J] + dp [I-1] [K], the node where both K and J are located cannot end with an invalid string.

On this basis, the high precision is achieved!

 

[Code]

#include <iostream>using namespace std;struct node{int index;node *fail;node *next[50];int count;}trie[1000];int total;node *root;node *q[1000];int head, tail;char word[100];struct num{int number[101];int len;};num dp[100][100];void add(num a, num &t){int i;for (i=0; i<a.len && i<t.len; i++)t.number[i] += a.number[i];if (a.len>t.len){for (i=t.len; i<a.len; i++){t.number[i] = a.number[i];}t.len = a.len;}for (i=1; i<t.len; i++){t.number[i] += t.number[i-1]/10;t.number[i-1] %= 10;}while (t.number[t.len-1]>10){t.number[t.len] = t.number[t.len-1]/10;t.number[t.len-1] %= 10;t.len++;}}int rank[300];int ct;int n,m,p;inline node *new_node(){node *p = &trie[total];p->index = total;p->count = 0;p->fail = NULL;int i;for (i=0; i<n; i++)p->next[i] = NULL;total++;return p;}void insert(char *s){node *p = root;int index;int i = 0;while(s[i]!='\0'){index = rank[s[i]];if (p->next[index]==NULL)p->next[index] = new_node();p = p->next[index];i++;}p->count++;}void build_ac_automation(){int i;node *temp;head = 1;tail = 0;q[0] = root;while(head!=tail){temp = q[tail];tail++;for (i=0; i<n; i++){if (temp->next[i]!=NULL){if (temp==root)temp->next[i]->fail = root;else{temp->next[i]->fail = temp->fail->next[i];if (temp->next[i]->fail->count!=0)temp->next[i]->count = 1;}q[head] = temp->next[i];head++;}else{if (temp==root)temp->next[i] = root;elsetemp->next[i] = temp->fail->next[i];}}}}num solve(){int i,j,k;int index;for (i=0; i<=m; i++){for (j=0; j<=total; j++){dp[i][j].len = 1;}}dp[0][0].len = 1;dp[0][0].number[0] = 1;for (i=1; i<=m; i++){for (j=0; j<total; j++){if (trie[j].count==0){for (k=0; k<n; k++){if (trie[j].next[k]->count==0){index = trie[j].next[k]->index;add(dp[i-1][j], dp[i][index]);}}}}}num sum;sum.len = 1;sum.number[0] = 0;for (i=0; i<total; i++){if (trie[i].count==0){add(dp[m][i], sum);}}return sum;}int main(){int i;scanf("%d %d %d", &n, &m, &p);scanf("%s", word);for (i=0,ct=0; word[i]!='\0'; i++){rank[word[i]] = ct;ct++;}total = 0;root = new_node();for (i=0; i<p; i++){scanf("%s", word);insert(word);}build_ac_automation();num t = solve();for (i=t.len-1; i>=0; i--)printf("%d", t.number[i]);printf("\n");return 0;}

[P.s]

TC has not been done yet.

The sgu may be refreshed later.