Codeforces 455B A Lot of Games (Dictionary tree + game), codeforces455b

Codeforces 455B A Lot of Games (Dictionary tree + game), codeforces455b

Question connection: Codeforces 455B A Lot of Games

Given n represents a string set. Given k, the game is played k times, followed by n strings. Starting from each board, the string is an empty string, and then the two append characters at the end in turn to ensure that the new string is the prefix of a string in the set and cannot be input by the operator, the new round begins with the previous sentence.

Solution: first, build a dictionary tree for the character set, and then search for the first-hand decision-making status based on the game's winning and losing nature. You can decide to win 3, only win 2, and only fail 1, uncontrollable (that is, the opponent can decide to win) 0.
For the status 3, to win, you can use the former K-1 field defeat, and then ensure that the K field first hand, take the winning solution.
For status 2, if both of them win, it must be that two people take turns to win, so K's parity is determined.
For statuses 1 and 0, it is necessary to lose because you keep losing and keep starting first.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e5+5;struct node {    int val, arr[30];    node () {        val = 0;        memset(arr, 0, sizeof(arr));    }}t[maxn*2];int top = 1, len, n, k;char str[maxn];inline int get_node () {    return top++;}void insert (int x, int d) {    if (d == len)        return;    int mv = str[d] - 'a';    if (t[x].arr[mv] == 0)        t[x].arr[mv] = get_node();    insert(t[x].arr[mv], d+1);}int solve (int x) {    int& ans = t[x].val;    ans = 0;    bool flag = true;    for (int i = 0; i < 26; i++) {        if (t[x].arr[i]) {            flag = false;            ans |= solve(t[x].arr[i]);        }    }    if (flag)        ans = 1;    return 3-ans;}int main () {    scanf("%d%d", &n, &k);    for (int i = 0; i < n; i++) {        scanf("%s", str);        len = strlen(str);        insert(0, 0);    }    solve(0);    int ans = t[0].val;    if (ans < 2)        printf("Second\n");    else if (ans == 2)        printf("%s\n", k&1 ? "First" : "Second");    else        printf("First\n");    return 0;}