HDU 4930 Fighting the Landlords(暴力枚舉+類比)

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HDU 4930 Fighting the Landlords

題目連結

題意：就是題中那幾種牌型，如果先手能一步走完，或者一步讓後手無法管上，就贏

思路：先枚舉出兩個人所有可能的牌型的最大值，然後再去判斷即可

代碼：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct Player {    int rank[15];} p1, p2;int t, hash[205], cnt[(1<<20)], vis[20];char a[25], b[25];int bitcount(int x) {    int ans = 0;    while (x) {ans += (x&1);x >>= 1;    }    return ans;}void add(int num, Player &p) {    if (num == 2) {if (vis[14] && vis[15]) {    p.rank[7] = 14;    return;}    }    if (num == 5) {int a = 0, b = 0;for (int i = 15; i >= 1; i--) {    if (vis[i] == 3) a = i;    if (vis[i] == 2) b = i;}if (a && b) {    p.rank[5] = max(p.rank[5], a);}return;    }    for (int i = 15; i >= 1; i--) {if (num == 4 && vis[i] == 4) {    p.rank[7] = max(p.rank[7], i);    return;}if (num == 4 && vis[i] == 3) {    p.rank[4] = max(p.rank[4], i);    return;}if (num == 6 && vis[i] == 4) {    p.rank[6] = max(p.rank[6], i);    return;}if (vis[i] == num) {    p.rank[num] = max(p.rank[num], i);    return;}    }}void build(char *a, Player &p) {    memset(p.rank, 0, sizeof(p.rank));    int n = strlen(a);    int maxs = (1<<n);    memset(vis, 0, sizeof(vis));    for (int i = 0; i < maxs; i++) {if (cnt[i] > 6) continue;memset(vis, 0, sizeof(vis));for (int j = 0; j < n; j++) {    if (i&(1<<j)) {vis[hash[a[j]]]++;    }}add(cnt[i], p);    }}bool solve() {    int n = strlen(a);    if (n == 4) {if (p1.rank[7]) return true;    }    if (n <= 6) {if (p1.rank[n]) return true;    }    if (p1.rank[7] && p2.rank[7]) return p1.rank[7] > p2.rank[7];    if (p1.rank[7] && !p2.rank[7]) return true;    if (!p1.rank[7] && p2.rank[7]) return false;    for (int i = 1; i < 7; i++) {if (p1.rank[i] > p2.rank[i]) return true;    }    return false;}int main() {    for (int i = 0; i < (1<<20); i++)cnt[i] = bitcount(i);    for (int i = 3; i <= 9; i++)hash[i + '0'] = i - 2;    hash['T'] = 8; hash['J'] = 9; hash['Q'] = 10; hash['K'] = 11;    hash['A'] = 12; hash['2'] = 13; hash['X'] = 14; hash['Y'] = 15;    scanf("%d", &t);    while (t--) {scanf("%s%s", a, b);build(a, p1);build(b, p2);if (solve()) printf("Yes\n");else printf("No\n");    }    return 0;}