POJ2945 Find the clones

传送门

Trie树的模板题。我们把每个人都存在Trie树里面，然后再每个人跑一遍即可。注意一个人走过之后要清零，否则的话会被重复计算。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')

using namespace std;
typedef long long ll;
const int M = 40005;
const int N = 500005;
const ll mod = 1000000007;

int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
    if(ch == '-') op = -1;
    ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
    ans *= 10;
    ans += ch - '0';
    ch = getchar();
    }
    return ans * op;
}

int n,m,ans[M];
char s[M][50];

struct trie
{
    int c[N][26],cnt,val[N];
    bool vis[N];
    void clear()
    {
        memset(c,0,sizeof(c));
        memset(val,0,sizeof(val));
        memset(vis,0,sizeof(vis));
        cnt = 0;
    }
    void insert(char *p)
    {
        int l = strlen(p),u = 0;
        rep(i,0,l-1)
        {
        int v = p[i] - 'A';
        if(!c[u][v]) c[u][v] = ++cnt;
        u = c[u][v];
        if(i == l-1) val[u]++;
        }
    }
    void find(char *p)
    {
        int l = strlen(p),u = 0;
        rep(i,0,l-1)
        {
        int v = p[i] - 'A';
        u = c[u][v];
        if(i == l-1 && val[u]) ans[val[u]]++,val[u] = 0;
        }
    }
}T;

int main()
{
    while(1)
    {
    n = read(),m = read();
    if(!n && !m) break;
    T.clear();
    memset(ans,0,sizeof(ans));
    rep(i,1,n) scanf("%s",s[i]),T.insert(s[i]);
    rep(i,1,n) T.find(s[i]);
    rep(i,1,n) printf("%d\n",ans[i]);
    }
    return 0;
}