LightOJ 1224 - DNA Prefix(字典树)

LightOJ 1224 - DNA Prefix

题目大意：给定若干个字符串，找到一个前缀，前缀长度*出现次数最大值。

解题思路：对字符串集合建立字典树，然后遍历一遍，每个节点等于dep*val。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 50005 * 50;
const int sigma_size = 4;

struct Tire {
    int sz, ans;
    int g[maxn][sigma_size];
    int val[maxn];

    void init();
    int idx(char ch);
    void insert(char* s);
    void solve(int u, int d);
}T;

int main () {
    int cas;
    scanf("%d", &cas);

    int n;
    char w[105];
    for (int kcas = 1; kcas <= cas; kcas++) {
        T.init();

        scanf("%d", &n);
        for (int i = 0; i < n; i++) {
            scanf("%s", w);
            T.insert(w);
        }

        T.solve(0, 0);
        printf("Case %d: %d\n", kcas, T.ans);
    }
    return 0;
}

void Tire::init() {
    sz = 1;
    ans = val[0] = 0;
    memset(g[0], 0, sizeof(g[0]));
}

int Tire::idx (char ch) {
    if (ch == 'A') return 0;
    else if (ch == 'C') return 1;
    else if (ch == 'G') return 2;
    else return 3;
}

void Tire::solve(int u, int d) {
    ans = max(ans, d * val[u]);

    for (int i = 0; i < 4; i++) {
        if (g[u][i])
            solve(g[u][i], d + 1);
    }
}

void Tire::insert(char* s) {
    int u = 0, n = strlen(s);

    for (int i = 0; i < n; i++) {
        int v = idx(s[i]);

        if (g[u][v] == 0) {
            val[sz] = 0;
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }

        u = g[u][v];
        val[u]++;
    }
}