hdu 1247 Hat’s Words

http://acm.hdu.edu.cn/showproblem.php?pid=1247



因为是两个词拼接，因此我们可以建两棵前缀树 遍历时分别记录true的位置看是否匹配。MLE无数发！也是醉了！

AC代码：

#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxnode = 100010;

vector <string> ans;
string str[maxnode];
int cnt1,cnt2,ans1[105],ans2[105],a,b;

struct Trie{
    bool flag;
    int next[30];
}zt[maxnode],ft[maxnode];

void Init(){
    cnt1 = cnt2 = 1;
    for(int i=0; i<maxnode; i++){
        memset(zt[i].next, -1, sizeof(zt[i].next));
        memset(ft[i].next, -1, sizeof(ft[i].next));
        zt[i].flag = false;
        ft[i].flag = false;
    }
}

void Insert1(string s){
    int p = 0, n = s.length();
    for(int i=0; i<n; i++){
        if(zt[p].next[s[i]-'a'] == -1)
            zt[p].next[s[i]-'a'] = cnt1++;
        p = zt[p].next[s[i]-'a'];
    }
    zt[p].flag = true;
}

void Insert2(string s){
    int p = 0, n = s.length();
    for(int i=n-1; i>=0; i--){
        if(ft[p].next[s[i]-'a'] == -1)
            ft[p].next[s[i]-'a'] = cnt2++;
        p = ft[p].next[s[i]-'a'];
    }
    ft[p].flag = true;
}

void Query1(string s){
    a = 0;
    memset(ans1, 0, sizeof(ans1));
    int p = 0, n = s.length(), tmp = 0;
    for(int i=0; i<n; i++){
        if(zt[p].flag)  if(tmp < n) ans1[a++] = tmp;
        p = zt[p].next[s[i]-'a'];
        tmp++;
    }
}

void Query2(string s){
    b = 0;
    memset(ans2, 0, sizeof(ans2));
    int p = 0, n = s.length(), tmp = 0;
    for(int i=n-1; i>=0; i--){
        if(ft[p].flag)  if(tmp < n) ans2[b++] = tmp;
        p = ft[p].next[s[i]-'a'];
        tmp++;
    }
}

int main(){
    //freopen("in.txt","r", stdin);
    Init();
    string tmp;
    int num = 0;
    while(cin >> tmp){
        str[num] = tmp;
        num++;
    }
    for(int i=0; i<num; i++){
        Insert1(str[i]);
        Insert2(str[i]);
    }
    for(int i=0; i<num; i++){
        Query1(str[i]);
        Query2(str[i]);
        bool flag = false;
        for(int u=0; u<a; u++){
            if(ans1[u]){
                int key = str[i].length() - ans1[u];
                for(int v=0; v<b; v++){
                    if(ans2[v] == key){
                        ans.push_back(str[i]);
                        flag = true;
                        break;
                    }
                }
            }
            if(flag)    break;
        }
    }
//    sort(ans.begin(), ans.end());
    for(vector <string>::iterator iter=ans.begin(); iter!=ans.end(); iter++)
        cout << *iter << endl;
    return 0;
}