hdu-1277--字典树坑题

Posted 2021-02-15 shuizhidao

hdu-1227

字典树，坑题！！当字典树练手

Problem Description

我们大家经常用google检索信息，但是检索信息的程序是很困难编写的；现在请你编写一个简单的全文检索程序。
问题的描述是这样的：给定一个信息流文件，信息完全有数字组成，数字个数不超过60000个，但也不少于60个；再给定一个关键字集合，其中关键字个数不超过10000个，每个关键字的信息数字不超过60个，但也不少于5个；两个不同的关键字的前4个数字是不相同的；由于流文件太长，已经把它分成多行；请你编写一个程序检索出有那些关键字在文件中出现过。

Input

第一行是两个整数M，N；M表示数字信息的行数，N表示关键字的个数；接着是M行信息数字，然后是一个空行；再接着是N行关键字；每个关键字的形式是：[Key No. 1] 84336606737854833158。

Output

输出只有一行，如果检索到有关键字出现，则依次输出，但不能重复，中间有空格，形式如：Found key: [Key No. 9] [Key No. 5]；如果没找到，则输出形如：No key can be found !。

Sample Input

20 10
646371829920732613433350295911348731863560763634906583816269
637943246892596447991938395877747771811648872332524287543417
420073458038799863383943942530626367011418831418830378814827
679789991249141417051280978492595526784382732523080941390128
848936060512743730770176538411912533308591624872304820548423
057714962038959390276719431970894771269272915078424294911604
285668850536322870175463184619212279227080486085232196545993
274120348544992476883699966392847818898765000210113407285843
826588950728649155284642040381621412034311030525211673826615
398392584951483398200573382259746978916038978673319211750951
759887080899375947416778162964542298155439321112519055818097
642777682095251801728347934613082147096788006630252328830397
651057159088107635467760822355648170303701893489665828841446
069075452303785944262412169703756833446978261465128188378490
310770144518810438159567647733036073099159346768788307780542
503526691711872185060586699672220882332373316019934540754940
773329948050821544112511169610221737386427076709247489217919
035158663949436676762790541915664544880091332011868983231199
331629190771638894322709719381139120258155869538381417179544
000361739177065479939154438487026200359760114591903421347697

[Key No. 1] 934134543994403697353070375063
[Key No. 2] 261985859328131064098820791211
[Key No. 3] 306654944587896551585198958148
[Key No. 4] 338705582224622197932744664740
[Key No. 5] 619212279227080486085232196545
[Key No. 6] 333721611669515948347341113196
[Key No. 7] 558413268297940936497001402385
[Key No. 8] 212078302886403292548019629313
[Key No. 9] 877747771811648872332524287543
[Key No. 10] 488616113330539801137218227609

Sample Output

Found key: [Key No. 9] [Key No. 5]

坑位：文件流字符串长度是6e5才能过！没有No key can be found!的数据！（算了，写都写了）

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <sstream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iomanip>
#include <stack>

using namespace std;

typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 60005;
const int MOD = 1e9 + 9;

#define lson l, m, rt << 14
#define rson m + 1, r, rt << 1 | 1

string s;
int tree[N * 10][15], vis[N * 10], pos;
char t[N];
set<int> p;
vector<int> v;
set<int>::iterator pit;
vector<int>::iterator vit;

void build(string s1, int num)
{
    int rt = 0;
    for(int i = 0;s1[i];++i)
    {
        int x = s1[i] - '0';
        if(!tree[rt][x])
            tree[rt][x] = ++pos;
        rt = tree[rt][x];
    }
    vis[rt] = num;
}

void solve(string s1)
{
    int rt = 0;

    for(int i = 0;s1[i];++i)
    {
        int x = s1[i] - '0';
        if(vis[rt])
        {
           if(p.find(vis[rt]) == p.end())
           {
               v.push_back(vis[rt]);
               p.insert(vis[rt]);
           }
           return;
        }
        if(!tree[rt][x]) return;
        rt = tree[rt][x];
    }
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 0;i < n;++i)
    {
        scanf("%s", t);
        s += t;
    }
    getchar();
    for(int i = 1;i <= m;++i)
    {
        int a;
        scanf("
[Key No. %d] %s", &a, t);
        build(t, i);
    }
    for(int i = 0;s[i];++i)
        solve(&s[0] + i);
    if(p.size())
    {
        printf("Found key:");
        for(vit = v.begin();vit != v.end();++vit)
            printf(" [Key No. %d]", *vit);
    }
    else
        printf("No key can be found!");
    printf("
");
    return 0;
}