codeforces132C

题意：有个乌龟在一条线上走，初始位置为1

给你一串FT字符串代表指令，F代表朝着当前前进的方向走一步，T代表转向，再给你转换次数n，

每次可以把T变F，F变T，同一个指令可以变多次

问你在转换了n次的情况下，走的最远距离（假设初始位置为0，最远左边为o则答案为abs(o - 0)）

很恶心的dp

dp[i][j][k][l] 已经执行了i指令，已经改变了j次，当前位置为k，朝向为l是否合法

最后遍历一遍k, l搜最远的k就行了

//
//  Created by Matrix on 2016-01-22
//  Copyright (c) 2015 Matrix. All rights reserved.
//
//
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <sstream>
#include <set>
#include <vector>
#include <stack>
#define ALL(x) x.begin(), x.end()
#define INS(x) inserter(x, x,begin())
#define ll long long
#define CLR(x) memset(x, 0, sizeof x)
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 1e6 + 10;
const int maxv = 1e3 + 10;
const double eps = 1e-9;

char buf[maxv];
int n;
bool dp[105][55][305][2];
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
while(scanf("%s", buf + 1) != EOF) { //1代表正方向，0代表反方向
int len = strlen(buf + 1);
scanf("%d", &n);
CLR(dp);
//      puts(buf+1);
dp[0][0][150][0] = true;
for(int i = 1; i <= len; i++) {
for(int j = 0; j <= n; j++) {
for(int k = 1; k <= 299; k++) {
if(buf[i] == 'T') {
for(int l = j; l >= 0; l -= 2)
dp[i][j][k][1] |= dp[i-1][l][k][0];
for(int l = j - 1; l >= 0; l -= 2)
dp[i][j][k][1] |= dp[i-1][l][k-1][1];

for(int l = j; l >= 0; l -= 2)
dp[i][j][k][0] |= dp[i-1][l][k][1];
for(int l = j - 1; l >= 0; l -= 2)
dp[i][j][k][0] |= dp[i-1][l][k+1][0];
}
else {
for(int l = j; l >= 0; l -= 2)
dp[i][j][k][1] |= dp[i-1][l][k-1][1];
for(int l = j - 1; l >= 0; l -= 2)
dp[i][j][k][1] |= dp[i-1][l][k][0];

for(int l = j; l >= 0; l -= 2)
dp[i][j][k][0] |= dp[i-1][l][k+1][0];
for(int l = j - 1; l >= 0; l -= 2)
dp[i][j][k][0] |= dp[i-1][l][k][1];
}
}
}
}
int ans = 0;
for(int k = 1; k <= 299; k++) {
for(int l = 0; l <= 1; l++) {
if(dp[len]
[k][l]) {
//                  printf("%d %d\n", k, l);
ans = max(ans, abs(k - 150));
}
}
}
printf("%d\n", ans);
}
return 0;
}