贪心+模拟 ZOJ 3829 Known Notation

 1 /*
 2     题意：一串字符串，问要最少操作数使得成为合法的后缀表达式
 3     贪心+模拟：数字个数 >= *个数+1 所以若数字少了先补上在前面，然后把不合法的*和最后的数字交换，记录次数
 4             岛娘的代码实在难懂啊~
 5 */
 6 /************************************************
 7 * Author        :Running_Time
 8 * Created Time  :2015-8-16 14:29:49
 9 * File Name     :K.cpp
10  ************************************************/
11 
12 #include <cstdio>
13 #include <algorithm>
14 #include <iostream>
15 #include <sstream>
16 #include <cstring>
17 #include <cmath>
18 #include <string>
19 #include <vector>
20 #include <queue>
21 #include <deque>
22 #include <stack>
23 #include <list>
24 #include <map>
25 #include <set>
26 #include <bitset>
27 #include <cstdlib>
28 #include <ctime>
29 using namespace std;
30 
31 #define lson l, mid, rt << 1
32 #define rson mid + 1, r, rt << 1 | 1
33 typedef long long ll;
34 const int MAXN = 1e3 + 10;
35 const int INF = 0x3f3f3f3f;
36 const int MOD = 1e9 + 7;
37 char str[MAXN];
38 
39 int main(void)    {     //ZOJ 3829 Known Notation
40     int T;  scanf ("%d", &T);
41     while (T--) {
42         scanf ("%s", str);
43         int x = 0, len = strlen (str);
44         for (int i=0; i<len; ++i)  {
45             if (str[i] == '*')  x++;
46         }
47         if (x == 0) {
48             puts ("0"); continue;
49         }
50         int ans = max (x + 1 - (len - x), 0);   int n = ans;
51         for (int i=0; i<len; ++i)   {
52             if (str[i] == '*')  {
53                 if (n <= 1) n++, ans++;
54                 else    n--;
55             }
56             else    n++;
57         }
58         printf ("%d
", ans);
59     }
60 
61     return 0;
62 }
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