hdu2222 Keywords Search

最新推荐文章于 2021-05-18 10:56:21 发布

原创最新推荐文章于 2021-05-18 10:56:21 发布 · 827 阅读

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CC 4.0 BY-SA版权

文章标签：

#search #tree #image #up #integer #ini

本文介绍了一种基于关键词搜索的图像检索系统实现方法，包括输入解析、关键词匹配及输出展示等关键步骤。

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17672 Accepted Submission(s): 5908

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

Input

First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.

Output

Print how many keywords are contained in the description.

Sample Input

1
5
she
he
say
shr
her
yasherhs

Sample Output

Author

Wiskey

Recommend

lcy

AC自动机第一题，自己理解写的，所以比较挫，不过纪念下，这个代码改了三天- -||||

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;

typedef struct
{
    int word,fail,p,num;
    int next[26];
}Tree;

Tree tree[15000005];
char str[1000005];
int up,ans;
queue <int> q;

void Add(char *str)
{
    int i,j,n,now;
    n=strlen(str);
    now=0;
    for (i=0;i<n;i++)
    {
        if (tree[now].next[str[i]-'a']!=-1) now=tree[now].next[str[i]-'a'];
        else
        {
            tree[now].next[str[i]-'a']=up;
            tree[up].word=tree[up].fail=0;
            tree[up].num=str[i]-'a';
            tree[up].p=now;
            memset(tree[up].next,-1,sizeof(tree[up].next));
            now=up++;
        }
    }
    tree[now].word++;
}

void BFS()
{
    int t,tag,i;
    q.push(0);
    while(!q.empty())
    {
        tag=q.front();
        q.pop();
        for (i=0;i<26;i++)
        {
            if (tree[tag].next[i]!=-1) q.push(tree[tag].next[i]);
        }
        if (tree[tag].p==0)
        {
            tree[tag].fail=0;
            continue;
        }
        t=tree[tree[tag].p].fail;
        while(1)
        {
            if (tree[t].next[tree[tag].num]==-1)
            {
                if (t==0)
                {
                    tree[tag].fail=0;
                    break;
                }
                t=tree[t].p;
            }
            else
            {
                tree[tag].fail=tree[t].next[tree[tag].num];
                break;
            }
        }
    }
}

void Count()
{
    int i,j,now,n,tag;
    now=0;
    n=strlen(str);
    for (i=0;i<n;i++)
    {
        while(tree[now].next[str[i]-'a']==-1 && now!=0)
        {
            now=tree[now].fail;
        }
        if (tree[now].next[str[i]-'a']!=-1) now=tree[now].next[str[i]-'a'];
        tag=now;
        while(tag!=0 && tree[tag].word!=-1)
        {
            ans+=tree[tag].word;
            tree[tag].word=-1;
            tag=tree[tag].fail;
        }
    }
}

int main()
{
    int T,i,j,n,now;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        tree[0].word=tree[0].fail=tree[0].p=0;
        memset(tree[0].next,-1,sizeof(tree[0].next));
        up=1;
        for (i=0;i<n;i++)
        {
            scanf("%s",str);
            Add(str);
        }
        BFS();
        scanf("%s",str);
        n=strlen(str);
        now=0;
        ans=0;
        Count();
        printf("%d\n",ans);
    }
    return 0;
}