牛客练习赛13E 乌龟跑步

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define rep(i,n) for (int i = 0; i < (n); ++i)
 5 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 6 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 7 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 8 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 9 
10 #define pr(x) cout << #x << " = " << x << "  "
11 #define prln(x) cout << #x << " = " << x << endl
12 
13 #define LOWBIT(x) ((x)&(-x))
14 
15 #define ALL(x) x.begin(),x.end()
16 #define INS(x) inserter(x,x.begin())
17 
18 #define ms0(a) memset(a,0,sizeof(a))
19 #define msI(a) memset(a,inf,sizeof(a))
20 #define msM(a) memset(a,-1,sizeof(a))
21 
22 inline int gc(){
23     static const int BUF = 1e7;
24     static char buf[BUF], *bg = buf + BUF, *ed = bg;
25     
26     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
27     return *bg++;
28 } 
29 
30 inline int ri(){
31     int x = 0, f = 1, c = gc();
32     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
33     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
34     return x*f;
35 }
36 
37 typedef long long LL;
38 typedef unsigned long long uLL;
39 const double EPS = 1e-9;
40 const int inf = 1e9 + 9;
41 const LL mod = 1e9 + 7;
42 const int maxN = 1e5 + 7;
43 const LL ONE = 1;
44 
45 
46 int n, m, ans; 
47 string c; 
48 // dp[i][j][k][0/1]表示已进行了i次指令，已修改了j次指令，当前走到k位置，朝向为前(后)这种情况能否到达 
49 // dp[i][j][k][0/1] = 1表示能到达，dp[i][j][k][0/1] = 0表示不能 
50 // 为避免负数问题，k初始值为100 
51 int dp[107][57][207][2];
52 
53 inline void dfs(int x, int y, int z, int d) {
54     if(x > m || y > n || dp[x][y][z][d]) return;
55     dp[x][y][z][d] = 1;
56     
57     if(c[x] == 'F') {
58         // 修改
59         dfs(x + 1, y + 1, z, !d); 
60         // 不修改 
61         dfs(x + 1, y, z + (d ? -1 : 1), d); 
62     }
63     if(c[x] == 'T') {
64         // 修改
65         dfs(x + 1, y + 1, z + (d ? -1 : 1), d); 
66         // 不修改 
67         dfs(x + 1, y, z, !d); 
68     }
69 }
70 
71 int main(){
72     cin >> c >> n;
73     m = c.size();
74     
75     dfs(0, 0, 100, 0);
76     
77     rep(k, 201) {
78         if(dp[m][n][k][0] || dp[m][n][k][1]) ans = max(ans, abs(k - 100));
79     }
80     
81     cout << ans << endl;
82     return 0;
83 }