LightOJ 1224 - DNA Prefix(字典树)

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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]++;    }}

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