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HDU 5384 Danganronpa(AC自动机)

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Danganronpa

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 429    Accepted Submission(s): 248


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)=2f(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 nm.
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.

T10
For each test case, n,m1051|Ai|,|Bj|104|Ai|105|Bj|105
For all test case, |Ai|6?105|Bj|6?105Ai 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
 

     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;
}



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HDU 5384 Danganronpa(AC自动机)

原文:http://blog.csdn.net/acm_baihuzi/article/details/47674317

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