牛客练习赛13E 乌龟跑步

题目链接：https://ac.nowcoder.com/acm/contest/70/E

题目大意：

　　略

分析：

　　DP或记忆化搜索，个人觉得记忆化搜索比较好做，逻辑清晰，代码量少

代码如下：

 #include <bits/stdc++.h>
 using namespace std;

 #define rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 const double EPS = 1e-;
 const int inf = 1e9 + ;
 const LL mod = 1e9 + ;
 const int maxN = 1e5 + ;
 const LL ONE = ;

 int n, m, ans;
 string c;
 // 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
 // dp[i][j][k][0/1] = 1表示能到达，dp[i][j][k][0/1] = 0表示不能
 // 为避免负数问题，k初始值为100
 int dp[][][][];

 inline void dfs(int x, int y, int z, int d) {
     if(x > m || y > n || dp[x][y][z][d]) return;
     dp[x][y][z][d] = ;

     if(c[x] == 'F') {
         // 修改
         dfs(x + , y + , z, !d);
         // 不修改
         dfs(x + , y, z + (d ? - : ), d);
     }
     if(c[x] == 'T') {
         // 修改
         dfs(x + , y + , z + (d ? - : ), d);
         // 不修改
         dfs(x + , y, z, !d);
     }
 }

 int main(){
     cin >> c >> n;
     m = c.size();

     dfs(, , , );

     rep(k, ) {
         if(dp[m][n][k][] || dp[m][n][k][]) ans = max(ans, abs(k - ));
     }

     cout << ans << endl;
     return ;
 }