Danganronpa

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 241 Accepted Submission(s): 137


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

1
5 6
orz
sto
kirigiri
danganronpa
ooooo
o
kyouko
dangan
ronpa
ooooo
ooooo

Sample Output

1
1
0
3
7

Source 2015 Multi-University Training Contest 8

#include 
using namespace std;
#define prt(k) cerr<<#k = < q;
    for (int i=0;i<26;i++) {
        int &v = ch[0][i];
        if (v != -1) {
            q.push(v);
            fail[v] = 0;
        } else v = 0;
    }
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i=0;i<26;i++) {
            int &v = ch[u][i];
            if (v==-1) {
                v = ch[fail[u]][i];
            } else {
                fail[v] = ch[fail[u]][i];
                q.push(v);
            }
        }
    }
}
ll query(char buf[], int len)
{
    ll ret = 0; int u = 0;
    for (int i=0;i0; v=fail[v])
            ret += ed[v];
    }
    return ret;
}
int n, m;
int main()
{
    int re, ca= 1; scanf(%d, &re);
    while (re--) {
        scanf(%d%d, &n, &m);
        pos[0] = 0;
        for (int i=0;i