hdu 5384 Danganronpa(AC自动机)

Danganronpa

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1154    Accepted Submission(s): 594

Problem Description

Danganronpa is a video game franchise created and developed by Spike Chunsoft, the series' name is compounded from the Japanese words for "bullet" (dangan) and "refutation" (ronpa).

Now, Stilwell is playing this game. There are n verbal evidences, and Stilwell has m "bullets". Stilwell will use these bullets to shoot every verbal evidence.

Verbal evidences will be described as some strings Ai, and bullets are some strings Bj. The damage to verbal evidence Ai from the bullet Bj is f(Ai,Bj).

f(A,B)=∑i=1|A|−|B|+1[ A[i...i+|B|−1]=B ]

In other words, f(A,B) is equal to the times that string B appears as a substring in string A.
For example: f(ababa,ab)=2, f(ccccc,cc)=4

Stilwell wants to calculate the total damage of each verbal evidence Ai after shooting all m bullets Bj, in other words is ∑mj=1f(Ai,Bj).

 

Input

The first line of the input contains a single number T, the number of test cases.
For each test case, the first line contains two integers n, m.
Next n lines, each line contains a string Ai, describing a verbal evidence.
Next m lines, each line contains a string Bj, describing a bullet.

T≤10
For each test case, n,m≤105, 1≤|Ai|,|Bj|≤104, ∑|Ai|≤105, ∑|Bj|≤105
For all test case, ∑|Ai|≤6∗105, ∑|Bj|≤6∗105, Ai and Bj consist of only lowercase English letters

 

Output

For each test case, output n lines, each line contains a integer describing the total damage of Ai from all m bullets, ∑mj=1f(Ai,Bj).

 

Sample Input

15 6orzstokirigiridanganronpaooooookyoukodanganronpaoooooooooo

 

Sample Output

11037
#include <stdio.h>#include <algorithm>#include <iostream>#include <string.h>#include <queue>#include<vector>#include<string>using namespace std;const int maxn = 1e5*5 + 200;struct Trie{    int next[maxn][26], fail[maxn], end[maxn],ans[maxn];    int root, L;    int newnode()    {        for (int i = 0; i < 26; i++)            next[L][i] = -1;        end[L++] = 0;        return L - 1;    }    void init()    {        L = 0;        memset(ans, 0, sizeof(ans));        root = newnode();    }        void insert(char buf[])    {        int len = strlen(buf);        int now = root;        for (int i = 0; i < len; i++)        {            if (next[now][buf[i] - 'a'] == -1)                next[now][buf[i] - 'a'] = newnode();            now = next[now][buf[i] - 'a'];        }        end[now]++;    }    void build()    {        queue<int>Q;        fail[root] = root;        for (int i = 0; i < 26; i++)            if (next[root][i] == -1)                next[root][i] = root;            else            {                fail[next[root][i]] = root;                Q.push(next[root][i]);            }        while (!Q.empty())        {            int now = Q.front();            Q.pop();            for (int i = 0; i < 26; i++)                if (next[now][i] == -1)                    next[now][i] = next[fail[now]][i];                else                {                    fail[next[now][i]] = next[fail[now]][i];                    Q.push(next[now][i]);                }        }    }    void query(string buf,int id)    {        int len = buf.size();        int now = root;        for (int i = 0; i < len; i++)        {            now = next[now][buf[i] - 'a'];            int temp = now;            while (temp != root)            {                ans[id] += end[temp];                temp = fail[temp];            }        }    }};Trie ac;vector<string>que;char buf[maxn];string s;int main(){    int T;    int n,m;    scanf("%d", &T);    while (T--)    {        scanf("%d%d", &n,&m);        ac.init();        que.clear();        for (int i = 0; i < n; i++)        {            cin >> s;            que.push_back(s);        }        for (int i = 0; i < m; i++)        {            scanf("%s", buf);            ac.insert(buf);        }        ac.build();        for (int i = 0; i < n; i++)            ac.query(que[i], i);        for (int i = 0; i < n; i++)            printf("%d\n", ac.ans[i]);    }    return 0;}