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Family View

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2099 Accepted Submission(s): 441

Problem Description Steam is a digital distribution platform developed by Valve Corporation offering digital rights management (DRM), multiplayer gaming and social networking services. A family view can help you to prevent your children access to some content which are not suitable for them.

Take an MMORPG game as an example, given a sentence T, and a list of forbidden words {P}, your job is to use ‘*‘ to subsititute all the characters, which is a part of the substring matched with at least one forbidden word in the list (case-insensitive).

For example, T is: "I love Beijing‘s Tiananmen, the sun rises over Tiananmen. Our great leader Chairman Mao, he leades us marching on."

And {P} is: {"tiananmen", "eat"}

The result should be: "I love Beijing‘s *********, the sun rises over *********. Our gr*** leader Chairman Mao, he leades us marching on."

Input The first line contains the number of test cases. For each test case:
The first line contains an integer n, represneting the size of the forbidden words list P. Each line of the next n lines contains a forbidden words Pi (1≤|Pi|≤1000000,∑|Pi|≤1000000) where Pi only contains lowercase letters.

The last line contains a string T (|T|≤1000000)

.

Output For each case output the sentence in a line.

Sample Input 1 3 trump ri o Donald John Trump (born June 14, 1946) is an American businessman, television personality, author, politician, and the Republican Party nominee for President of the United States in the 2016 election. He is chairman of The Trump Organization, which is the principal holding company for his real estate ventures and other business interests.

Sample Output D*nald J*hn ***** (b*rn June 14, 1946) is an Ame**can businessman, televisi*n pers*nality, auth*r, p*litician, and the Republican Party n*minee f*r President *f the United States in the 2016 electi*n. He is chairman *f The ***** *rganizati*n, which is the p**ncipal h*lding c*mpany f*r his real estate ventures and *ther business interests. 分析： 題意讓我們把列表中的出現的單詞，在文章中全部替換為*（無論大小寫） 多元匹配的問題，我們可以想到用AC自動機. 我們可以用end[j]表示以j節點結尾的單詞的最大長度， 那麽如果匹配到當前的第j個字符且end[j]>0,將從後往前的end[j]長度的字符串全部化為* 大小問題的處理方式.我們可以先標記大寫字母，然後將文章全部化為小寫字母， 這樣的話，在輸出的時候，如果該字母沒有被變成*,且被標記，就輸出對應的大寫字母 代碼如下：

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
int vis[1000010];
struct Trie
{
    int Next[1000010][26];//26是這裏討論26個小寫字母的情況，根據情況修改
    int fail[1000010],end[1000010];//end數組表示以該節點結尾的字符串的數量
    int len[1000010];
    int root,L;//L用來標記節點序號，以廣度優先展開的字典樹的序號
    int newnode()  //建立新節點
    {
        for(int i = 0;i < 26;i++)
            Next[L][i] = -1;     //將該節點的後繼節點域初始化
        len[L]=0;
        end[L++] = 0;
        return L-1;    //返回當前節點編號
    }
    void init() //初始化操作
    {
        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)  //如果未建立當前的後繼節點，建立新的節點
                Next[now][buf[i]-‘a‘] = newnode();
                len[Next[now][buf[i]-‘a‘]]=len[now]+1;
            now = Next[now][buf[i]-‘a‘];
        }
        end[now]=len[now];//以該節點結尾的字符串數量增加1
    }
    void build()
    {
        queue<int>Q; //用廣度優先的方式，將樹層層展開
        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();
            end[now]=max(end[now],end[fail[now]]);
            for(int i = 0;i < 26;i++)
                if(Next[now][i] == -1)
                    Next[now][i] = Next[fail[now]][i];//該段的最後一個節點匹配後，跳到擁有最大公共後綴的fail節點繼續匹配
                else
                {
                    fail[Next[now][i]]=Next[fail[now]][i];//當前節點的fail節點等於它前驅節點的fail節點的後繼節點
                    Q.push(Next[now][i]);
                }
        }
    }
    void solve(char buf[])
    {
        int L=strlen(buf);
        int now=root;
        for(int i=0;i<L;i++)
        {
          now=Next[now][buf[i]-‘a‘];
          if(end[now]>0)
          {
             for(int j=i;j>i-end[now];j--)
              buf[j]=‘*‘;
          }
        }
        for(int i=0;i<L;i++)
        {
            if(buf[i]!=‘*‘&&vis[i]==1)
            printf("%c",buf[i]-32);
            else
            printf("%c",buf[i]);
        }
        printf("\n");
    }
};
char buf[1000010];
Trie ac;
int main()
{
    int  t,n,ans;
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        ac.init();
       scanf("%d",&n);
       for(int i=0;i<n;i++)
       {
           scanf("%s",buf);
           ac.insert(buf);
       }
       ac.build();
         getchar();
         gets(buf);
         for(int i=0;i<strlen(buf);i++)
         {
            if(buf[i]>=‘A‘&&buf[i]<=‘Z‘){
             buf[i]=buf[i]+32;
             vis[i]=1;
            }
         }
         ac.solve(buf);
    }
    return 0;
}

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
int vis[1000010];
struct Trie
{
int Next[1000010][26];//26是這裏討論26個小寫字母的情況，根據情況修改
int fail[1000010],end[1000010];//end數組表示以該節點結尾的字符串的數量
int len[1000010];
int root,L;//L用來標記節點序號，以廣度優先展開的字典樹的序號
int newnode() //建立新節點
{
for(int i = 0;i < 26;i++)
Next[L][i] = -1; //將該節點的後繼節點域初始化
len[L]=0;
end[L++] = 0;
return L-1; //返回當前節點編號
}
void init() //初始化操作
{
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) //如果未建立當前的後繼節點，建立新的節點
Next[now][buf[i]-‘a‘] = newnode();
len[Next[now][buf[i]-‘a‘]]=len[now]+1;
now = Next[now][buf[i]-‘a‘];
}
end[now]=len[now];//以該節點結尾的字符串數量增加1
}
void build()
{
queue<int>Q; //用廣度優先的方式，將樹層層展開
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();
end[now]=max(end[now],end[fail[now]]);
for(int i = 0;i < 26;i++)
if(Next[now][i] == -1)
Next[now][i] = Next[fail[now]][i];//該段的最後一個節點匹配後，跳到擁有最大公共後綴的fail節點繼續匹配
else
{
fail[Next[now][i]]=Next[fail[now]][i];//當前節點的fail節點等於它前驅節點的fail節點的後繼節點
Q.push(Next[now][i]);
}
}
}
void solve(char buf[])
{
int L=strlen(buf);
int now=root;
for(int i=0;i<L;i++)
{
now=Next[now][buf[i]-‘a‘];
if(end[now]>0)
{
for(int j=i;j>i-end[now];j--)
buf[j]=‘*‘;
}
}
for(int i=0;i<L;i++)
{
if(buf[i]!=‘*‘&&vis[i]==1)
printf("%c",buf[i]-32);
else
printf("%c",buf[i]);
}
printf("\n");
}
};
char buf[1000010];
Trie ac;
int main()
{
int t,n,ans;
scanf("%d",&t);
while(t--)
{
memset(vis,0,sizeof(vis));
ac.init();
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%s",buf);
ac.insert(buf);
}
ac.build();
getchar();
gets(buf);
for(int i=0;i<strlen(buf);i++)
{
if(buf[i]>=‘A‘&&buf[i]<=‘Z‘){
buf[i]=buf[i]+32;
vis[i]=1;
}
}
ac.solve(buf);
}
return 0;
}

HDU 5880 Family View （AC自動機）