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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;i
0; 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
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