ZOJ 2852 Deck of Cards(DP)

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题目连接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1852

题意太繁琐就不讲了.

设dp[q][i][j][k]表示前i张牌,每个卡槽的分值各为i,j,k时候的最能的多金钱数.

dp[0][0][0][0]=0,其他设为-1.

用了滚动数组优化空间,具体的转移方程看程序吧.

#include <cstdio>#include <algorithm>#include <memory.h>#define JOKER 21#define FACE 10using namespace std;const int MAX = 101;int convert_to_val(char c){if(c == 'F')return JOKER;else if(c == 'A')return 1;else if(c >= '2' && c <= '9')return c - '0';else if(c == 'T' || c == 'J' || c == 'Q' || c == 'K')return FACE;}int main(int argc, char const *argv[]){int dp[2][31][31][31];int N;while(scanf("%d", &N) && N){int ans = -1;memset(dp, -1, sizeof(dp));dp[0][0][0][0] = 0;int f = 0;for(int p = 0; p <= N; ++p, f ^= 1){int nf = f ^ 1;char card[2];if(p != N)scanf("%s", card);elsecard[0] = 'F';int val = convert_to_val(card[0]);for(int i = 0; i < 31; ++i){for(int j = 0; j < 31; ++j){for(int k = 0; k < 31; ++k){if(dp[f][i][j][k] == -1)continue;ans = max(ans, dp[f][i][j][k]);if(val == JOKER){//joker caseif(i <= 20)dp[nf][0][j][k] = max(dp[nf][0][j][k], dp[f][i][j][k] + 50 + 100);if(j <= 20)dp[nf][i][0][k] = max(dp[nf][i][0][k], dp[f][i][j][k] + 50 + 200);if(k <= 20)dp[nf][i][j][0] = max(dp[nf][i][j][0], dp[f][i][j][k] + 50 + 300);}else{int nv = i + val;if(nv == 21){//reach 21dp[nf][0][j][k] = max(dp[nf][0][j][k], dp[f][i][j][k] + 50 + 100);}else if(i < 21){dp[nf][nv][j][k] = max(dp[nf][nv][j][k], dp[f][i][j][k] + 50);}nv = j + val;if(nv == 21){//reach 21dp[nf][i][0][k] = max(dp[nf][i][0][k], dp[f][i][j][k] + 50 + 200);}else if(j < 21){dp[nf][i][nv][k] = max(dp[nf][i][nv][k], dp[f][i][j][k] + 50);}nv = k + val;if(nv == 21){//reach 21dp[nf][i][j][0] = max(dp[nf][i][j][0], dp[f][i][j][k] + 50 + 300);}else if(k < 21){dp[nf][i][j][nv] = max(dp[nf][i][j][nv], dp[f][i][j][k] + 50);}}}}}memset(dp[f], -1, sizeof(dp[f]));}printf("%d\n", ans);}return 0;}

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