题意：斗地主，就是要自己出牌。使得对手在这一轮无法出牌，或者有出牌的可能。可是你的牌已经走完了。假设符合这些条件的话，输出Yes。否则输出No。
思路：先预处理能直接把牌走完的情况，假设不行的话就直接暴力枚举能获胜的情况。
#include#include #include #include using namespace std;const int MAXN = 20;char s1[MAXN], s2[MAXN];int p1[MAXN], p2[MAXN], num1[MAXN], num2[MAXN];void change(char ch, int *p) { if (ch == 'Y') p[17] = 1; else if (ch == 'X') p[16] = 1; else if (ch == '2') p[15]++; else if (ch == 'A') p[14]++; else if (ch == 'K') p[13]++; else if (ch == 'Q') p[12]++; else if (ch == 'J') p[11]++; else if (ch == 'T') p[10]++; else p[ch - '0']++; }void count(int n, int *num) { if (n == 1) num[1]++; else if (n == 2) num[2]++; else if (n == 3) num[3]++; else if (n == 4) num[4]++;}int slove(int n) { if (n == 1) return 1; else if (n == 2 && num1[2]) return 1; else if (n == 2 && p1[16] && p1[18]) return 1; else if (n == 3 && num1[3]) return 1; else if (n == 4 && (num1[4] || (num1[3] && num1[1]))) return 1; else if (n == 5 && num1[3] && num1[2]) return 1; else if (n == 6 && num1[4]) return 1; return 0;}int main() { int cas; scanf("%d", &cas); while (cas--) { scanf("%s%s", s1, s2); int l1 = strlen(s1); int l2 = strlen(s2); memset(p1, 0, sizeof(p1)); memset(p2, 0, sizeof(p2)); for (int i = 0; i < l1; i++) change(s1[i], p1); for (int i = 0; i < l2; i++) change(s2[i], p2); memset(num1, 0, sizeof(num1)); memset(num2, 0, sizeof(num2)); for (int i = 0; i < 18; i++) count(p1[i], num1); for (int i = 0; i < 18; i++) count(p2[i], num2); int flag = 0; if (l1 <= 6) //推断是否能直接把牌走完的情况 flag = slove(l1); if (flag) { printf("Yes\n"); continue; } if (p1[16] && p1[17]) { //推断两方是否存在一方有王炸的情况 printf("Yes\n"); continue; } if (p2[16] && p2[17]) { printf("No\n"); continue; } if (num1[4]) { //推断炸的存在的情况 int a = 1, b = 1; for (int i = 18; i >= 3; i--) if (p1[i] == 4) { a = i; break; } for (int i = 18; i >= 3; i--) if (p2[i] == 4) { b = i; break; } if (a > b) flag = 1; if (b > a) flag = -1; } if (flag == 1) { printf("Yes\n"); continue; } if (flag == -1) { printf("No\n"); continue; } int c = 0; //推断之存在对手有炸的情况 for (int i = 18; i >= 3; i--) if (p2[i] == 4) { c = 1; break; } if (c) { printf("No\n"); continue; } if (num1[3]) { //推断有三带是否能直接获胜的情况 int a = 1, b = 1; for (int i = 18; i >= 3; i--) if (p1[i] == 3) { a = i; break; } for (int i = 18; i >= 3; i--) if (p2[i] == 3) { b = i; break; } if (num1[2] && !num2[2]) { flag = 1; } if (a > b) flag = 1; } if (num1[2]) { //推断出对子是否能直接获胜的情况 int a = 1, b = 1; for (int i = 18; i >= 3; i--) if (p1[i] == 2 || p1[i] == 3) { a = i; break; } for (int i = 18; i >= 3; i--) if (p2[i] == 2 || p2[i] == 3) { b = i; break; } if (a >= b) flag = 1; } if (num1[1]) { //推断出单是否能直接获胜的情况 int a = 1, b = 1; for (int i = 18; i >= 3; i--) if (p1[i]) { a = i; break; } for (int i = 18; i >= 3; i--) if (p2[i]) { b = i; break; } if (a >= b) flag = 1; } if (flag) printf("Yes\n"); else printf("No\n"); } return 0;}