Deck of Cards ZOJ - 2852 dp 多决策 三维 滚动更新

题意：一个特殊21点游戏 具体http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2852

题解：建一个三维dp，表示三个卡槽分别为i,j,l分时最大的收益情况。

　　　　对所有当前状态dp，将下一个可能的状态存入f，

坑：~-1==0

#define _CRT_SECURE_NO_WARNINGS
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<iomanip>
#include<cmath>
#include<cstdio>
#include<string>
#include<stack>
#include<ctime>
#include<list>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<sstream>
#include<fstream>
#include<iostream>
#include<functional>
#include<algorithm>
#include<memory.h>
//#define INF 0x3f3f3f3f
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mp make_pair
#define pb push_back
#define mmm(a,b) memset(a,b,sizeof(a))
//std::ios::sync_with_stdio(false);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
void smain();
#define ONLINE_JUDGE
int main() {
    //ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    FILE *myfile;
    myfile =freopen("C:\\Users\\SuuTT\\Desktop\\test\\in.txt", "r", stdin);
    if (myfile == NULL)
        fprintf(stdout, "error on input freopen\n");
    /*FILE *outfile;
    outfile= freopen("C:\\Users\\SuuTT\\Desktop\\test\\out.txt", "w", stdout);
    if (outfile == NULL)
        fprintf(stdout, "error on output freopen\n");*/
    long _begin_time = clock();
#endif
    smain();
#ifndef ONLINE_JUDGE
    long _end_time = clock();
    printf("time = %ld ms.", _end_time - _begin_time);
#endif
    return ;
}
int dir[][] = { ,,,,-,,,- };
const int maxn = 2e2 + ;
int f[][][], dp[][][];

int n; int ans;
int num(char c) {
    if (c >= ''&&c <= '')return c - '';
    else if (c == 'A')return ;
    else if (c == 'F')return -;
    else return ;
}

void smain() {
    while (cin >> n) {
        if (n == )break;
        mmm(f, ); mmm(dp, );
        string c;
        cin >> c;
        int x = num(c[]);
        if (x==-) {
            //cout << ~x << endl;
            dp[][][] = ;
            ans = ;
        }
        else {
            dp[x][][] = ;
            dp[][x][] = ;
            dp[][][x] = ;
            ans = ;
        }
        rep(i, , n) {
            string c;
            cin >> c;
            int x = num(c[]);
            rep(j, , )rep(k, , )rep(l, , ) if(dp[j][k][l]){
                int t = dp[j][k][l];
                if ((x == - && j < ) || x + j == ) f[][k][l] = max(f[][k][l], t + );//放第一组//正好21，[j][k][l]->[0][k][l]
                else if (x + j < )f[j + x][k][l] = max(f[x + j][k][l], t + );//没到21[j][k][l]->[x + j][k][l]
                else if (x + j >  && j < )f[][k][l] = max(f[][k][l], t + );//超了21[j][k][l]->[21][k][l]

                if ((x == - && k < ) || x + k == ) { f[j][][l] = max(f[j][][l], t + ); }
                else if (x + k < )f[j][k+x][l] = max(f[j][k+x][l], t + );
                else if (x + k >  && k < )f[j][][l] = max(f[j][][l], t + );

                if ((x == - && l < ) || x + l == ) { f[j][k][] = max(f[j][k][], t + ); }
                else if (x + l < )f[j][k][l+x] = max(f[j][k][l+x], t + );
                else if (x + l >  && l < )f[j][k][] = max(f[j][k][], t + );
            }
            rep(j, , )rep(k, , )rep(l, , )dp[j][k][l] = f[j][k][l], ans = max(ans,dp[j][k][l]), f[j][k][l] = ;
        }
        cout << ans << endl;
    }

}