hdu 1528 Card Game Cheater 二分图的经典应用
2014-07-29 17:24
363 查看
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
[align=left]Sample Output[/align]
[align=left]Source[/align]
Northwestern Europe 2004
可以理解 为:把小于第二行的数记为联通,则最多有多少条,就有多少point;
#include<stdio.h>
#include<string.h>
#define maxn 100
bool Map[maxn][maxn];
bool Flag[maxn];
int Mark[maxn];
int lx[maxn];
int ly[maxn];
bool Find(int x,int n)
{
int i;
for(i=1;i<=n;i++)
{
if(!Map[x][i]||Flag[i])
continue;
Flag[i]=true;
if(!Mark[i]||Find(Mark[i],n))
{
Mark[i]=x;
return true;
}
}
return false;
}
int change(char *str)
{
int t=0,k;
int i;
for(i=0;i<=1;i++)
{
t*=10;
if(str[i]=='2')
k=2;
if(str[i]=='3')
k=3;
if(str[i]=='4')
k=4;
if(str[i]=='5')
k=5;
if(str[i]=='6')
k=6;
if(str[i]=='7')
k=7;
if(str[i]=='8')
k=8;
if(str[i]=='9')
k=9;
if(str[i]=='T')
k=10;
if(str[i]=='J')
k=11;
if(str[i]=='Q')
k=12;
if(str[i]=='K')
k=13;
if(str[i]=='A')
k=14;
if(str[i]=='H')
k=4;
if(str[i]=='S')
k=3;
if(str[i]=='D')
k=2;
if(str[i]=='C')
k=1;
t+=k;
}
return t;
}
int main()
{
char str[5];
int n,m;
int i,j;
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
for(i=1;i<=m;i++)
{
scanf("%s",str);
lx[i]=change(str);
}
for(i=1;i<=m;i++)
{
scanf("%s",str);
ly[i]=change(str);
}
for(i=1;i<=m;i++)
for(j=1;j<=m;j++)
if(lx[i]<ly[j])
Map[i][j]=true;
else
Map[i][j]=false;
int Count=0;
memset(Mark,false,sizeof(Mark));
for(i=1;i<=m;i++)
{
memset(Flag,false,sizeof(Flag));
if(Find(i,m))
Count++;
}
printf("%d\n",Count);
}
return 0;
}
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.
[align=left]Input[/align]
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
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
3 1 JD JH 2 5D TC 4C 5H 3 2H 3H 4H 2D 3D 4D
[align=left]Sample Output[/align]
1 1 2
[align=left]Source[/align]
Northwestern Europe 2004
可以理解 为:把小于第二行的数记为联通,则最多有多少条,就有多少point;
#include<stdio.h>
#include<string.h>
#define maxn 100
bool Map[maxn][maxn];
bool Flag[maxn];
int Mark[maxn];
int lx[maxn];
int ly[maxn];
bool Find(int x,int n)
{
int i;
for(i=1;i<=n;i++)
{
if(!Map[x][i]||Flag[i])
continue;
Flag[i]=true;
if(!Mark[i]||Find(Mark[i],n))
{
Mark[i]=x;
return true;
}
}
return false;
}
int change(char *str)
{
int t=0,k;
int i;
for(i=0;i<=1;i++)
{
t*=10;
if(str[i]=='2')
k=2;
if(str[i]=='3')
k=3;
if(str[i]=='4')
k=4;
if(str[i]=='5')
k=5;
if(str[i]=='6')
k=6;
if(str[i]=='7')
k=7;
if(str[i]=='8')
k=8;
if(str[i]=='9')
k=9;
if(str[i]=='T')
k=10;
if(str[i]=='J')
k=11;
if(str[i]=='Q')
k=12;
if(str[i]=='K')
k=13;
if(str[i]=='A')
k=14;
if(str[i]=='H')
k=4;
if(str[i]=='S')
k=3;
if(str[i]=='D')
k=2;
if(str[i]=='C')
k=1;
t+=k;
}
return t;
}
int main()
{
char str[5];
int n,m;
int i,j;
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
for(i=1;i<=m;i++)
{
scanf("%s",str);
lx[i]=change(str);
}
for(i=1;i<=m;i++)
{
scanf("%s",str);
ly[i]=change(str);
}
for(i=1;i<=m;i++)
for(j=1;j<=m;j++)
if(lx[i]<ly[j])
Map[i][j]=true;
else
Map[i][j]=false;
int Count=0;
memset(Mark,false,sizeof(Mark));
for(i=1;i<=m;i++)
{
memset(Flag,false,sizeof(Flag));
if(Find(i,m))
Count++;
}
printf("%d\n",Count);
}
return 0;
}
相关文章推荐
- HDU 1528 Card Game Cheater(匈牙利算法,二分图最大匹配):
- HDU1528_Card Game Cheater(二分图/最大匹配)
- HDU 1528 Card Game Cheater(二分图最大匹配)
- HDU 1528 Card Game Cheater 二分图最大匹配
- hdu 1528 Card Game Cheater(二分图的最大匹配,匈牙利算法)
- hdu 1528 Card Game Cheater(二分图的最大匹配)
- HDU 1528 Card Game Cheater(二分图最大匹配)
- HDU 1528 Card Game Cheater (二分图求最大匹配)
- 杭电 hdu 1528 Card Game Cheater (二分图,最大匹配)
- hdu 1528 Card Game Cheater(二分图最大匹配)
- hdu----(1528)Card Game Cheater(最大匹配/贪心)
- hdu 1528 Card Game Cheater ( 二分图匹配 )
- POJ 2062 & HDOJ 1528 Card Game Cheater - 阅读理解..二分图最大匹配
- HDOJ 题目1528 Card Game Cheater(二分图最小点覆盖)
- hdu 1528 Card Game Cheater (最小覆盖)
- hdu 1528 Card Game Cheater
- hdu 1528-Card Game Cheater(贪心算法)
- HDU-1528/1962 Card Game Cheater
- hdu 1528 Card Game Cheater
- hdu 1528【Card Game Cheater】