HDU5384 Danganronpa

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

 

ac自动机模板题啊，直接上模板

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<queue>#include<vector>using namespace std;typedef long long ll;const int size = 26;//字典树节点大小const int base = 'a';//字典树const int maxn = 100005;//字典树大小（便于各种操作）class tire{public:    tire *next[size], *fail;    ll cnt, id;    //各种节点设置    tire(){ cnt = 0; fail = NULL; memset(next, 0, sizeof(next)); };};//字典树节点设置class Ac_machine{private:    tire *root;//根节点建立    tire *node[maxn];//与id相对立，便于各种操作    int tot;//字典树大小    char s[maxn];//读入字符串public:    char S[maxn];//母串    Ac_machine(){}    void clear(){ node[0] = root = new tire; tot = 0; }//初始化    void insert(int);//插入多个字符串    int insert();//插入字符串,返回字符串大小    void getfail();    //获取失配指针    int getmother();//获得匹配的母串，返回母串大小    int work_out();//求解函数，依照题目不同而不同,返回答案}ac;//ac自动机设置int Ac_machine::getmother(){ scanf("%s", S); return strlen(S); }void Ac_machine::insert(int n){ while (n--) insert(); }int Ac_machine::insert(){    scanf("%s", s);    tire *now = root;    for (int i = 0, k; s[i]; i++)    {        k = s[i] - base;        if (!now->next[k])        {            node[++tot] = now->next[k] = new tire;            now->next[k]->id = tot;        }        now = now->next[k];    }    now->cnt++;    //可以插入一些字典树的设置    return strlen(s);}void Ac_machine::getfail(){    queue<tire*> p;    root->fail = root;    for (int i = 0; i < size; i++)    if (root->next[i])    {        root->next[i]->fail = root;        p.push(root->next[i]);    }    else root->next[i] = root;    tire *now;    while (!p.empty())    {        now = p.front();    p.pop();        now->cnt += now->fail->cnt;        //可以插入统计子串个数等操作        for (int i = 0; i < size; i++)        if (now->next[i])        {            now->next[i]->fail = now->fail->next[i];            p.push(now->next[i]);        }        else now->next[i] = now->fail->next[i];    }}int Ac_machine::work_out(){    ll ans = 0;    tire *now = root;    //for (int i = 0; i < maxn; i++) printf("%d\n", node[i]->cnt);    for (int i = 0, k; S[i]; i++)    {        k = S[i] - base;        now = now->next[k];        ans += now->cnt;    }    return ans;//返回子串出现次数}vector<char> p[maxn];char s[maxn];int main(){    int T,n,m;    scanf("%d",&T);    while (T--)    {        ac.clear();        scanf("%d%d",&n,&m);        for (int i=1;i<=n;i++)        {            p[i].clear();            scanf("%s",s);            for (int j=0;s[j];j++) p[i].push_back(s[j]);        }        ac.insert(m);        ac.getfail();        for (int i=1;i<=n;i++)        {            for (int j=0;j<p[i].size();j++)            {                ac.S[j]=p[i][j];            }            ac.S[p[i].size()]=0;            printf("%lld\n",ac.work_out());        }    }}