問題描述

Adam and Eve play a card game using a regular deck of 52 cards. The rules are simple. The players sit on opposite sides of a table, facing each other. Each player gets k cards from the deck and, after looking at them, places the cards face down in a row on the table. Adam’s cards are numbered from 1 to k from his left, and Eve’s cards are numbered 1 to k from her right (so Eve’s i:th card is opposite Adam’s i:th card). The cards are turned face up, and points are awarded as follows (for each i ∈ {1, . . . , k}):


If Adam’s i:th card beats Eve’s i:th card, then Adam gets one point.


If Eve’s i:th card beats Adam’s i:th card, then Eve gets one point.


A card with higher value always beats a card with a lower value: a three beats a two, a four beats a three and a two, etc. An ace beats every card except (possibly) another ace.


If the two i:th cards have the same value, then the suit determines who wins: hearts beats all other suits, spades beats all suits except hearts, diamond beats only clubs, and clubs does not beat any suit.

For example, the ten of spades beats the ten of diamonds but not the Jack of clubs.

This ought to be a game of chance, but lately Eve is winning most of the time, and the reason is that she has started to use marked cards. In other words, she knows which cards Adam has on the table before he turns them face up. Using this information she orders her own cards so that she gets as many points as possible.

Your task is to, given Adam’s and Eve’s cards, determine how many points Eve will get if she plays optimally.

輸入

There will be several test cases. The first line of input will contain a single positive integer N giving the number of test cases. After that line follow the test cases.

Each test case starts with a line with a single positive integer k <= 26 which is the number of cards each player gets. The next line describes the k cards Adam has placed on the table, left to right. The next line describes the k cards Eve has (but she has not yet placed them on the table). A card is described by two characters, the first one being its value (2, 3, 4, 5, 6, 7, 8 ,9, T, J, Q, K, or A), and the second one being its suit (C, D, S, or H). Cards are separated by white spaces. So if Adam’s cards are the ten of clubs, the two of hearts, and the Jack of diamonds, that could be described by the line

TC 2H JD

輸出

For each test case output a single line with the number of points Eve gets if she picks the optimal way to arrange her cards on the table.

樣例輸入

3 1 JD JH 2 5D TC 4C 5H 3 2H 3H 4H 2D 3D 4D

樣例輸出

1 1 2 題意：2個(gè)人打牌一張牌由2個(gè)字符表示 第一個(gè)是（23456789TJQKA） 第二個(gè)是（CDSH） 大小關(guān)系從左到右遞增? 現(xiàn)在第2個(gè)人知道了第一個(gè)人的牌 問在最好的打發(fā)下第二個(gè)人能贏幾次？ （貪心就搞就是待自己寫個(gè)排序。） #include<cstdio> #include<cstring> #include<stack> #include<vector> #include<queue> #include<cmath> #include<cstdlib> #include<iostream> #include<algorithm> using namespace std; typedef unsigned long long LL; typedef double de; const LL oo = 0x3f3f3f3f; const LL maxn = 1e6+7; const LL mod = 1e9+7; char str[20]= {"23456789TJQKA"}; char str1[20] = {"CDSH"}; struct da {char s[10]; } Adam[222], You[222]; void Sort(char s1[], char s2[])///把2個(gè)字符串升序排列 {char ss[20];int i, ai, bi;for(i = 0; i <= 12; i++){if(s1[0] == str[i]) ai = i;if(s2[0] == str[i]) bi = i;}if(ai < bi) return ;if(ai > bi){strcpy(ss, s1);strcpy(s1, s2);strcpy(s2, ss);return ;}for(i = 0; i < 4; i++){if(s1[1] == str1[i]) ai = i;if(s2[1] == str1[i]) bi = i;}if(ai > bi){strcpy(ss, s1);strcpy(s1, s2);strcpy(s2, ss);}return ; } int judge(char s1[], char s2[]) /// 返回1表示s1<s2 {int i;char a, b;int ai, bi;a = s1[0];b = s2[0];for(i = 0; i <= 12; i++){if(str[i] == a) ai = i;if(str[i] == b) bi = i;}if(ai < bi) return 1;if(ai > bi) return 0;a = s1[1];b = s2[1];ai = bi = -1;for(i = 0; i < 4; i++){if(str1[i] == a) ai = i;if(str1[i] == b) bi = i;}if(ai < bi) return 1;return 0; }int main() {int T, k, i, j, ans;scanf("%d", &T);while(T--){ans = 0;scanf("%d", &k);for(i = 0; i < k; i++)scanf("%s", Adam[i].s);for(i = 0; i < k; i++)scanf("%s", You[i].s);for(i = 0; i < k-1; i++){for(j = i+1; j < k; j++)Sort(Adam[i].s, Adam[j].s);}for(i = 0; i < k-1; i++){for(j = i+1; j < k; j++)Sort(You[i].s, You[j].s);}i = j = 0;///排完序從頭遍歷一邊就好 i，j分別指向2個(gè)人的第幾張牌while(i < k && j < k){if(judge(Adam[i].s, You[j].s) == 1){ans ++;i++;j++;}else j++;}printf("%d\n", ans);}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/PersistFaith/p/4948084.html

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