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Card Game Cheater
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1569 Accepted Submission(s): 825
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.
Input 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
Output 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.
Sample Input
3
1
JD
JH
2
5D TC
4C 5H
3
2H 3H 4H
2D 3D 4D
Sample Output
1
1
2
题意: 亚当和夏娃打牌,夏娃能看见亚当的牌,可以有策略的出牌。田忌赛马模型。
主要把这些牌的字符转化为数字,再模拟田忌赛马的贪心算法就行了。
代码如下:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[30],b[30];
int getnum(char c)
{
if(c=='T')
return 10;
else if(c=='J')
return 11;
else if(c=='Q')
return 12;
else if(c=='K')
return 13;
else if(c=='A')
return 14;
else
return c-'0';
}
int getcnt(char c)
{
if(c=='C')
return 1;
else if(c=='D')
return 2;
else if(c=='S')
return 3;
else
return 4;
}
int main()
{
int t,n,i,amin,amax,bmin,bmax,awin,bwin;
char s[4];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=0;i<n;++i)
{
scanf("%s",s);
a[i]=getnum(s[0])*10+getcnt(s[1]);//字符转化为数字
}
for(i=0;i<n;++i)
{
scanf("%s",s);
b[i]=getnum(s[0])*10+getcnt(s[1]);
}
sort(a,a+n);
sort(b,b+n);
awin=bwin=0;//记录两者赢的次数
amin=bmin=0;//记录两者最小的牌
amax=bmax=n-1;//记录两者最大的牌
for(i=0;i<n;++i)
{
if(a[amin]<b[bmin])
{
bwin++;
amin++; bmin++;
}
else if(a[amin]>b[bmin])//夏娃最小的牌比亚当的最小牌小,则消耗其最大牌
{
awin++;
amax--; bmin++;
}
else//两者最小牌相等的时候
{
if(b[bmax]>a[amax])
{
bwin++;
bmax--; amax--;
}
else //当夏娃最大牌比亚当最大牌小时,则用最小牌消耗其最大牌
{
if(b[bmin]<a[amax])
{
awin++;
amax--; bmin++;
}
}
}
}
printf("%d\n",bwin);
}
return 0;
}
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