POJ 2287 - Tian Ji -- The Horse Racing（貪心）

題意：田忌和齊王都有 n（n <= 1000）匹馬，兩者進行 n 輪賽馬，贏一局+200，平一局得 0，輸一局 -200，求田忌最多能贏多少錢。

貪心，先將田忌和齊王的馬分別排序，而後最上等與最上等比，最下等與最下等比，若是有能贏的則+200，若是沒法知足條件，則用田忌的最下等與齊王的最上等比（故意輸或者平局）。

代碼以下：

 1 #pragma comment(linker, "/STACK:102400000, 102400000")
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cctype>
 5 #include<cstdlib>
 6 #include<cmath>
 7 #include<iostream>
 8 #include<sstream>
 9 #include<iterator>
10 #include<algorithm>
11 #include<string>
12 #include<vector>
13 #include<set>
14 #include<map>
15 #include<deque>
16 #include<queue>
17 #include<stack>
18 #include<list>
19 #define fin freopen("in.txt", "r", stdin)
20 #define fout freopen("out.txt", "w", stdout)
21 #define pr(x) cout << #x << " : " << x << "   "
22 #define prln(x) cout << #x << " : " << x << endl
23 #define Min(a, b) a < b ? a : b
24 #define Max(a, b) a < b ? b : a
25 typedef long long ll;
26 typedef unsigned long long llu;
27 const int INT_INF = 0x3f3f3f3f;
28 const int INT_M_INF = 0x7f7f7f7f;
29 const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
30 const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
31 const double pi = acos(-1.0);
32 const double EPS = 1e-8;
33 const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
34 const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
35 const ll MOD = 1e9 + 7;
36 using namespace std;
37 
38 #define NDEBUG
39 #include<cassert>
40 const int MAXN = 1000 + 10;
41 const int MAXT = 10000 + 10;
42 
43 int n;
44 deque<int> t, k;
45 
46 int main(){
47     while(scanf("%d", &n) == 1 && n){
48         int tmp;
49         t.clear();
50         k.clear();
51         for(int i = 0; i < n; ++i)  {scanf("%d", &tmp);  t.push_back(tmp);}
52         for(int i = 0; i < n; ++i)  {scanf("%d", &tmp);  k.push_back(tmp);}
53         sort(t.begin(), t.end());
54         sort(k.begin(), k.end());
55         int ans = 0;
56         while(n--){
57             if(t.front() > k.front()){
58                 t.pop_front();
59                 k.pop_front();
60                 ans += 200;
61             }
62             else if(t.back() > k.back()){
63                 t.pop_back();
64                 k.pop_back();
65                 ans += 200;
66             }
67             else{
68                 if(t.front() != k.back())  ans -= 200;
69                 t.pop_front();
70                 k.pop_back();
71             }
72         }
73         printf("%d\n", ans);
74     }
75     return 0;
76 }