Danganronpa

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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

题意：给n个母串，给m个匹配串，求每个母串依次和匹配串匹配，能得到的数目和。

题解：AC自动机模板题

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
struct Trie {
    int next[500010][26],fail[500010],end[500010];
    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;
        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]);
                }
        }
    }
    int query(char buf[]) {
        int len = strlen(buf);
        int now = root;
        int res = 0;
        for(int i = 0; i < len; i++) {
            now = next[now][buf[i]-'a'];
            int temp = now;
            while( temp != root ) {
                res += end[temp];
                //end[temp] = 0;
                temp = fail[temp];
            }
        }
        return res;
    }
    void debug() {
        for(int i = 0; i < L; i++) {
            printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
            for(int j = 0; j < 26; j++)
                printf("%2d",next[i][j]);
            printf("]\n");
        }
    }
};
char buf[100010];
int m;
char s[605015];
char c[10010];
Trie ac;
int main() {
   // freopen("test.in","r",stdin);
    int T;
    int n;
    scanf("%d",&T);
    while( T-- ) {
        scanf("%d%d",&n,&m);
        ac.init();
        memset(s,0,sizeof s);
        int len=0;
        for(int i=0; i<n; i++) { ///把所有的串接起来，用空格隔开
            scanf("%s",c);
            int l=strlen(c);
            if(i!=0)s[len++]=' ';
            strcat(s,c);
            len+=l;
        }
        s[len]=' ';
        s[++len]=0;
        for(int i=0; i<m; i++) {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        int l=0;
        while(l<len) {
            int k=0;
            while(s[l]!=' ') {///找到当前单词
                buf[k++]=s[l++];
            }
            buf[k]=0;
            l++;
            printf("%d\n",ac.query(buf));
        }
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。