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NAIPC2018

参考：http://www.cnblogs.com/LQLlulu/p/9513669.html?tdsourcetag=s_pctim_aiomsg
https://www.cnblogs.com/clrs97/p/8730429.html?tdsourcetag=s_pctim_aiomsg

E-Pre?x Free Code

题目描述

Consider n initial strings of lower case letters, where no initial string is a prefix of any other initial string. Now, consider choosing k of the strings (no string more than once), and concatenating them together. You can make this many such composite strings:
n × (n ? 1) × (n ? 2) × . . . × (n ? k + 1)
Consider sorting all of the composite strings you can get via this process in alphabetical order. You are given a test composite string, which is guaranteed to belong on this list. find the position of this test composite string in the alphabetized list of all composite strings, modulo 109 +7. The first composite string in the list is at position 1.

输入

Each input will consist of a single test case. Note that your program may be run multiple times on different inputs. Each test case will begin with a line with two integers, first n and then k(1 ≤ k ≤ n), where n is the number of initial strings, and k is the number of initial strings you choose to form composite strings. The upper bounds of n and k are limited by the constraints on the strings, in the following paragraphs.
Each of the next n lines will contain a string, which will consist of one or more lower case letters a..z. These are the n initial strings. It is guaranteed that none of the initial strings will be a prefix of any other of the initial strings.
finally, the last line will contain another string, consisting of only lower case letters a..z. This is the test composite string, the position of which in the sorted list you must find. This test composite string is guaranteed to be a concatenation of k unique initial strings.
The sum of the lengths of all input strings, including the test string, will not exceed 106 letters.

输出

Output a single integer, which is the position in the list of sorted composite strings where the test composite string occurs. Output this number modulo 109 + 7.

样例输入

5 3
a
b
c
d
e
cad

样例输出

26

一开始的思路：

用map把输入的字符串按字典序标记一个数字，因为涉及到排序和map 就要用string。然后再将要处理的字符串也化为数字，然后就是求这个排列是第几个，对于每位数字，减去前面有多少比它小的-1，然后乘以后面的全排列就好了。终于找前面有多少个比它小的，当然是暴力啦1e6/2=1e3最后半小时才意识到要用树状数组，然而还是内存超限。

正解：

怎么没想到字典树呢?好吧其实是想到的但是不会先建树，因为题中说没有单词是另一个单词的前缀，cnt[]维护每个单词的编号。然后再用一个num[]维护一下单词的字典次序，这个怎么维护呢？直接跑一遍tire，就可以了。num[cnt[u]]相当于之前方法的字符串转化为数字：正好复习一下tire模板

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e6+5;
const int mod=1e9+7;
int n,k,ant;
int s[maxn],c[maxn],cnt[maxn],num[maxn];
int tire[maxn][30];
ll f[1000010],invf[1000010];
char str[maxn];
ll qmul(ll a,ll b){ll ret=0; while(b){ if(b&1) ret=(ret+a)%mod; a=(a+a)%mod; b>>=1;} return ret;}
ll qpow(ll a,ll b){ll ret=1; while(b){ if(b&1) ret=qmul(ret,a); a=qmul(a,a); b>>=1;} return ret;}
void init(){
  f[0]=1;
  for(ll i=1;i<=1000001;i++) f[i]=f[i-1]*i%mod;
  invf[1000001]=qpow(f[1000001],mod-2);
  for(ll i=1000000;i>=0;i--) invf[i]=invf[i+1]*(i+1)%mod;
}
inline int lowbit(int x){return x&-x;}
inline void add(int x,int val){
  for(int i=x;i<=n;i+=lowbit(i)){
    c[i]+=val;
  }
}
inline int getsum(int x){
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i)){
        ret+=c[i];
    }
    return ret;
}
void ins(char *str)
{
    int len=strlen(str);
    int p=0;
    for(int i=0;i<len;i++){
        int ch=str[i]-'a';
        if(!tire[p][ch])
            tire[p][ch]=++ant;
            p=tire[p][ch];
    }
    cnt[p]++;
    //cout<<p<<" "<<cnt[p]<<endl;
}
//int srh(char *str)
//{
//    int ans=0;
//    int len=strlen(str);
//    int p=0;
//    for(int i=0;i<len;i++){
//        int ch=str[i]-'a';
//        p=tire[p][ch];
//        if(!p) return ans;
//        ans+=cnt[p];
//    }
//    return ans;
//}
int op;
void cal(int u)
{

    if(cnt[u]){
        op++;
        num[u]=op;
        return;
    }
    for(int i=0;i<26;i++){
        if(tire[u][i])
            cal(tire[u][i]);
    }
}
int main()
{
    init();
    ios::sync_with_stdio(false);
    cin.tie(0);
    while(cin>>n>>k){
        ant=0;
        memset(cnt,0,sizeof(cnt));
        memset(tire,0,sizeof(tire));
        memset(c,0,sizeof(c));
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++){
            cin>>str;
            ins(str);
        }
        cin>>str;
        op=0;
        cal(0);
        ll ans=0;
        ll t;
          int x;
        int u=0,id=0;
        int len=strlen(str);
        for(int i=0;i<len;i++){
            t=str[i]-'a';
            u=tire[u][t];
            if(cnt[u]){
                id++;
                x=getsum(num[u]);
                ans=(ans+qmul(qmul((num[u]-1-x),f[n-id]),invf[n-k]))%mod;
                add(num[u],1);
                u=0;
            }
        }
        cout<<(ans+1)%mod<<endl;//%lxq
    }
    return 0;
}

比赛时疯狂瞎改，想尽了所有加快的方法，加了tire之后跑得超快，直接第一.....

EXAM-2018-08-26

标签：space   number   ESS   oid   constrain   lock   not   ide   tput   

原文地址：https://www.cnblogs.com/smallocean/p/9539437.html