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, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

typedef long long LL;
typedef unsigned long long uLL;
const double EPS = 1e-9;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;


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[107][57][207][2];

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] = 1;
    
    if(c[x] == 'F') {
        // 修改
        dfs(x + 1, y + 1, z, !d); 
        // 不修改 
        dfs(x + 1, y, z + (d ? -1 : 1), d); 
    }
    if(c[x] == 'T') {
        // 修改
        dfs(x + 1, y + 1, z + (d ? -1 : 1), d); 
        // 不修改 
        dfs(x + 1, y, z, !d); 
    }
}

int main(){
    cin >> c >> n;
    m = c.size();
    
    dfs(0, 0, 100, 0);
    
    rep(k, 201) {
        if(dp[m][n][k][0] || dp[m][n][k][1]) ans = max(ans, abs(k - 100));
    }
    
    cout << ans << endl;
    return 0;
}