题目大意：你手中有五张牌，桌上叠着五张牌，你有超能力，可以让任意张手中的牌与桌上的任意张交换，使得交换以后手中与桌上依旧五张牌。问你交换的最好结果为何种牌？

解题思路：又有一些细节不同。规则如下：

1.straight-flush：同花顺，牌面为T（10） - A，这里不论花色是否相同；

2.four-of-a-kind：四条，牌面有4个相同的值；

3.full-house：船牌，牌面有3个相同值，剩下2个也相同值；

4.flush：同花，五张牌的花色相同，不是同花顺；

5.straight：顺子，五张牌的值连续，A可以作为1也可以作为14；

6.three-of-a-kind：三条，牌面有3个相同的值；

7.two-pairs：两对，牌面有2个对子；

8.one-pair：一对，牌面有一个对子，即2个同值；

9.highest-card：大牌，没有以上牌型。

了解规则以后，用子集生成法的增量构造法，枚举所有手中保留的为原手牌的哪一些，取出桌面上的那些牌也确定了。然后就是判断比较麻烦一些。严格遵守规则就好。顺子的判断，我是用了一个数组存手上的牌的值，要注意不越界。

ac代码：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
char hand[5][5], desk[5][5], kind[10][50];
int re, a[5];
struct node{
  int value;
  int color;
}own[5];
void init()
{
  re = 8;
  strcpy(kind[0], "straight-flush");
  strcpy(kind[1], "four-of-a-kind");   
  strcpy(kind[2], "full-house");   
  strcpy(kind[3], "flush");   
  strcpy(kind[4], "straight");   
  strcpy(kind[5], "three-of-a-kind");   
  strcpy(kind[6], "two-pairs");   
  strcpy(kind[7], "one-pair");   
  strcpy(kind[8], "highest-card");       
}
int tran(char ch)
{
  if (ch == 'K')
    return 13;
  if (ch == 'Q')
    return 12;
  if (ch == 'J')
    return 11;
  if (ch == 'T')
    return 10;
  if (ch == 'A')
    return 1;
return ch-'0';
}
void hand_card(int n)
{
  for (int i=0; i<n; i++){
own[i].value = tran(hand[ a[i] ][0]); 
own[i].color = hand[ a[i] ][1];
}
for (int i=n,j=0; i<5; i++,j++){
    own[i].value = tran(desk[j][0]);
    own[i].color = desk[j][1];
  }
}
bool compare(node a, node b)
{
  if (a.value != b.value)
    return a.value < b.value;
return a.color < b.color;
}
void best()
{
  int three_kind, two_kind, num[14], temp;
  memset(num, 0, sizeof(num));
  three_kind = two_kind = 0;
  sort(own, own+5, compare);
  for (int i=0; i<5; i++)
num[own[i].value]++;
if (num[1] && num[10] && num[11] && 
num[12] && num[13])
re = 0;
for (int i=0; i<=13; i++){
    if (num[i] == 4 && re > 1)
      re = 1;
    else if (num[i] == 2)
      two_kind++;
    else if (num[i] == 3)
      three_kind++;
  }
  if (three_kind && two_kind && re > 2)
    re = 2;     
  if (own[0].color == own[1].color && 
  own[1].color == own[2].color &&
  own[2].color == own[3].color && 
  own[3].color == own[4].color && re > 3)
      re = 3;
  temp = own[0].value;
  if (num[temp%14] && num[(temp+1)%14] && num[(temp+2)%14] 
&& num[(temp+3)%14] && num[(temp+4)%14] && re > 4)
    re = 4;   
  if (three_kind && re > 5)
    re = 5;
  if (two_kind == 2 && re > 6)
    re = 6;
  if (two_kind && re > 7)
    re = 7;
}

void solve(int cur)
{
  int minu=0;
  hand_card(cur);
  best();
  if (cur)
    minu = a[cur-1] + 1;
  for (int i=minu; i<5; i++){
a[cur] = i;
solve(cur+1);
}
}

int main()
{
while (scanf("%s", &hand[0])!=EOF){
init();
for (int i=1; i<5; i++)
      scanf("%s", &hand[i]);
    for (int i=0; i<5; i++)
scanf("%s", &desk[i]);
printf("Hand: ");
for (int i=0; i<5; i++)
      printf("%s ", hand[i]);
    solve(0);
    printf("Deck: ");
    for (int i=0; i<5; i++)
printf("%s ", desk[i]);
printf("Best hand: %s\n", kind[re]);
}
return 0;
}