【DP】ZOJ 2852 Deck of Cards

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ZOJ 2852 Deck of Cards
参考自：AC_Von
f[j][k][l]表示slot x1, x2, x3的值为j，k，l时的最优值。pre[j][k][l]表示他的前一个状态。
#include <iostream>#include <algorithm>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <cmath>using namespace std;int f[30][30][30], p[30][30][30];int char2num(char c){    if(c>='2' && c<='9') return c-'0';    if(c == 'A') return 1;    if(c == 'F') return -1;    return 10;}int main(int argc, char const *argv[]){    // freopen("in", "r", stdin);    int n;    while(scanf("%d", &n), n){        memset(f, 0, sizeof(f));        memset(p, 0, sizeof(p));        char str[2];        int ans = 0, t, tt;        scanf("%s", str);        t = char2num(str[0]);        if(t == -1){            p[0][0][0] = 350;            ans = 350;        }else{            p[t][0][0] = 50;            p[0][t][0] = 50;            p[0][0][t] = 50;            ans = 0;        }        for(int i = 0; i < n-1; i++){            scanf("%s", str);            t = char2num(str[0]);            for(int j=0; j<22; j++){                for(int k=0; k<22; k++){                    for(int l=0; l<22; l++){                        if((tt = p[j][k][l]) == 0) continue;                        //1x                        if((t == -1 && j < 21) || t + j == 21) {                            f[0][k][l] = max(f[0][k][l], tt + 150);                        } else if(t + j < 21) {                            f[j+t][k][l] = max(f[t+j][k][l], tt + 50);                        } else if(t + j > 21 && j < 21) {                            f[21][k][l] = max(f[21][k][l], tt + 50);                        }                        //2x                        if((t == -1 && k < 21) || t + k == 21) {                            f[j][0][l] = max(f[j][0][l], tt + 250);                        } else if(t + k < 21) {                            f[j][t+k][l] = max(f[j][t+k][l], tt + 50);                        } else if(t + k > 21 && k < 21) {                            f[j][21][l] = max(f[j][21][l], tt + 50);                        }                        //3x                        if((t == -1 && l < 21) || t + l == 21) {                            f[j][k][0] = max(f[j][k][0], tt + 350);                        } else if(t + l < 21) {                            f[j][k][l+t] = max(f[j][k][l+t], tt + 50);                        } else if(t + l > 21 && l < 21) {                            f[j][k][21] = max(f[j][k][21], tt + 50);                        }                    }                }            }            for(int j=0; j<22; j++){                for(int k=0; k<22; k++){                    for(int l=0; l<22; l++){                        p[j][k][l] = f[j][k][l];                        ans = max(ans, p[j][k][l]);                        f[j][k][l] = 0;                    }                }            }        }        printf("%d\n", ans);    }    return 0;}
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