hdu 5384 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

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

 

Sample Output

1 1 0 3 7

 

Author

SXYZ

 

Source

2015 Multi-University Training Contest 8

 

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给出n个A串和m个B串，对于每个A串，输出里面出现了几次B串，ac自动机对所有B串构建字典树，然后进行匹配查找。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define MAX 100001
using namespace std;

int pos;
int trie[MAX][26],fail[MAX] = {-1},sum[MAX];
char s[100001][10001];
void Insert_Str(char *s) {///字符串插入到字典树中
   int i = -1,r = 0;
   while(s[++ i]) {
       int d = s[i] - 'a';
       if(!trie[r][d]) {
           trie[r][d] = ++ pos;
       }
       r = trie[r][d];
   }
   sum[r] ++;
}
void Build_Fail() {///通过父结点的Fail更新子结点的Fail
    queue<int> q;
    q.push(0);
    while(!q.empty()) {
        int id = q.front();
        q.pop();
        for(int i = 0;i < 26;i ++) {
            if(trie[id][i]) {///第i个儿子存在
                int temp = fail[id];///temp赋值当前节点的Fail
                while(temp != -1) {
                    if(trie[temp][i]) {
                        fail[trie[id][i]] = trie[temp][i];
                        break;
                    }
                    temp = fail[temp];
                }
                q.push(trie[id][i]);
            }
        }
    }
}
void Ac_automation(int k) {
    int i = -1,r = 0,ans = 0;
    while(s[k][++ i]) {
        int d = s[k][i] - 'a';
        while(r && !trie[r][d]) r = fail[r];///如果没有匹配的子结点 就找它的Fail看看有没有匹配的子结点
        if(trie[r][d]) r = trie[r][d];
        int temp = r;
        while(temp) {
            ans += sum[temp];
            temp = fail[temp];///找最长后缀
        }
    }
    printf("%d
",ans);
}
int main() {
    char str[10001];
    int t,n,m;
    scanf("%d",&t);
    while(t --) {
        pos = 0;
        memset(trie,0,sizeof(trie));
        memset(fail + 1,0,sizeof(fail));
        memset(sum,0,sizeof(sum));
        scanf("%d %d",&n,&m);
        for(int i = 0;i < n;i ++) {
            scanf("%s",s[i]);
        }
        for(int i = 0;i < m;i ++) {
            scanf("%s",str);
            Insert_Str(str);
        }
        Build_Fail();
        for(int i = 0;i < n;i ++) {
            Ac_automation(i);
        }
    }
    return 0;
}