题目：
Description
Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are different. They divide cards between them in some manner, it's possible that they have different number of cards. Then they play a "war"-like card game.
The rules are following. On each turn a fight happens. Each of them picks card from the top of his stack and puts on the table. The one whose card value is bigger wins this fight and takes both cards from the table to the bottom of his stack. More precisely, he first takes his opponent's card and puts to the bottom of his stack, and then he puts his card to the bottom of his stack. If after some turn one of the player's stack becomes empty, he loses and the other one wins.
You have to calculate how many fights will happen and who will win the game, or state that game won't end.
Input
First line contains a single integer n (2 ≤ n ≤ 10), the number of cards.
Second line contains integer k1 (1 ≤ k1 ≤ n - 1), the number of the first soldier's cards. Then follow k1 integers that are the values on the first soldier's cards, from top to bottom of his stack.
Third line contains integer k2 (k1 + k2 = n), the number of the second soldier's cards. Then follow k2 integers that are the values on the second soldier's cards, from top to bottom of his stack.
All card values are different.
Output
If somebody wins in this game, print 2 integers where the first one stands for the number of fights before end of game and the second one is 1 or 2 showing which player has won.
If the game won't end and will continue forever output - 1.
题目大意：一个K1张牌，一人K2张牌，两个人都取自己最上面那张牌比较，牌大的这两张牌按小上大下的顺序放入自己牌堆下方，直到一方没牌停止。如果循环则没结果。
解题思路：前一种不循环的直接用数组和循环，但判断循环的数组就不行了，先把牌存储到字符数组，再订个map的二维数组，用字符做下标映射整数，每次在循环里判断是否相等。
代码：
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include <string>
5 #include <map>
6 #include <set>
7 #include <algorithm>
8 using namespace std;
9 const int maxn=20+20;
10 map<string,map<string,int> > p;
11 int main()
12 {
13 int n;
14 int k1,k2;
15 int a[maxn],b[maxn],c[maxn];
16 char s1[maxn],s2[maxn];
17 int i,j,k;
18 scanf("%d",&n);
19 scanf("%d",&k1);
20 for(i = 0; i<k1; i++)
21 {
22 scanf("%d",&a[i]);
23 c[i] = a[i];
24 }
25 scanf("%d",&k2);
26 for(i = 0; i<k2; i++)
27 {
28 scanf("%d",&b[i]);
29 c[k1+i] = b[i];
30 }
31 sort(c,c+k1+k2);
32 for(i = 0; i<k1; i++)
33 {
34 for(j = 0; j<k1+k2; j++)
35 {
36 if(a[i]==c[j])
37 s1[i] = j+'0';
38 }
39 }
40 s1[k1] = ' ';
41 for(i = 0; i<k2; i++)
42 {
43 for(j = 0; j<k1+k2; j++)
44 {
45 if(b[i]==c[j])
46 s2[i] = j+'0';
47 }
48 }
49 s2[k2] = ' ';
50 p[s1][s2] = 1;
51 int o = 0;
52 while(k1&&k2)
53 {
54 int p1 = s1[0],p2 = s2[0];
55 if(p1>p2)
56 {
57 for(i = 0; i<k2; i++)
58 s2[i] = s2[i+1];
59 k2--;
60 for(i = 0; i<k1; i++)
61 s1[i] = s1[i+1];
62 s1[k1-1] = p2;
63 s1[k1] = p1;
64 k1++;
65 s2[k2] = s1[k1] = ' ';
66 }
67 else
68 {
69 for(i = 0; i<k1; i++)
70 s1[i] = s1[i+1];
71 k1--;
72 for(i = 0; i<k2; i++)
73 s2[i] = s2[i+1];
74 s2[k2-1] = p1;
75 s2[k2] = p2;
76 k2++;
77 s2[k2] = s1[k1] = ' ';
78 }
79 if(p[s1][s2])
80 {
81 o = -1;
82 break;
83 }
84 o++;
85 p [s1][s2] = 1;
86 }
87 printf("%d",o);
88 if(o!=-1)
89 {
90 if(k1)
91 printf(" 1");
92 else
93 printf(" 2");
94 }
95 printf("
");
96 return 0;
97 }