Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem A - Yahtzee

The game of Yahtzee involves 5 dice, which are thrown in 13 rounds. A score card contains 13 categories; each round may be scored in a category of the player's choosing, but each category may be scored only once in the game. The 13 categores are scored as follows:

• ones - sum of all ones thrown
• twos - sum of all twos thrown
• threes - sum of all threes thrown
• fours - sum of all fours thrown
• fives - sum of all fives thrown
• sixes - sum of all sixes thrown

• chance - sum of all dice

• three of a kind - sum of all dice, provided at least three have same value
• four of a kind - sum of all dice, provided at least four have same value
• five of a kind - 50 points, provided all five dice have same value
• short straight - 25 points, provided four of the dice form a sequence (that is, 1,2,3,4 or 2,3,4,5 or 3,4,5,6)
• long straight - 35 points, provided all dice form a sequence (1,2,3,4,5 or 2,3,4,5,6)
• full house - 40 points, provided three of the dice are equal and the other two dice are also equal. (for example, 2,2,5,5,5)

Each of the last six categories may be scored as 0 if the criteria are not met.

The score for the game is the sum of all 13 categories plus a bonus of 35 points if the sum of the first six categores (ones to sixes) is 63 or greater.

Your job is to compute the best possible score for a sequence of rounds.

The Input

Each line of input contains 5 integers between 1 and 6, indicating the values of the five dice thrown in each round. There are 13 such lines for each game, and there may be any number of games in the input data.

The Output

Your output should consist of a single line for each game containing 15 numbers: the score in each category (in the order given), the bonus score (0 or 35), and the total score. If there is more than categorization that yields the same total score, any one will do.

Sample Input

1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 1 1 1 1
6 6 6 6 6
6 6 6 1 1
1 1 1 2 2
1 1 1 2 3
1 2 3 4 5
1 2 3 4 6
6 1 2 6 6
1 4 5 5 5
5 5 5 5 6
4 4 4 5 6
3 1 3 6 3
2 2 2 4 6

Output for Sample Input

1 2 3 4 5 0 15 0 0 0 25 35 0 0 90
3 6 9 12 15 30 21 20 26 50 25 35 40 35 327

---

光是看懂計分方式就很浪費時間了。

"sum of all ones thrown" 這到底什麼意思呢？所有骰到點數 1 的總合，類推。

那這題要怎麼做呢？每種骰子只對應一種分類方式，組合就有 13!，又有特別的 bonus 分數計算，
當前六種分類方式總和 pt >= 63 額外加 35 分，問最高總分為多少？

為了加快速度，做法有點跟背包問題相似，逐步放入一骰子，然後進行狀態轉移。
在這裡必須將狀態壓縮，[0000000000010] // 代表第二個分類方式已經被使用。
然後記錄 dp[分類可使用的狀態][前六個分類的總分] = 最高分。

為什麼要多一個 [前六個分類的總分] 是因為到後面要計算 bonus 的時候要用的，
既然大於等於 63 都有 bonus，那麼記錄 0-63 個狀態就可以了。



#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
#define STATE 8192 //1<<13
#define STATE2 64 // bonus use
#define DSIZE 13

int DICE[13][5];
int score_cat(int dice[], int cat) {
    int tot = 0, i;
    int D[7];
    switch(cat) {
        case 1:case 2:case 3:case 4:case 5:case 6:
            for(i = 0; i < 5; i++)
                if(dice[i] == cat)
                    tot += cat;
            break;
        case 7: // chance
            for(i = 0; i < 5; i++)
                tot += dice[i];
            break;
        case 8: // three of a kind
            if(dice[0] == dice[2] || dice[1] == dice[3] ||
               dice[2] == dice[4])
            for(i = 0; i < 5; i++)
                tot += dice[i];
            break;
        case 9: // four of a kind
            if(dice[0] == dice[3] || dice[1] == dice[4])
            for(i = 0; i < 5; i++)
                tot += dice[i];
            break;
        case 10: // five of a kind
            if(dice[0] == dice[4])
                tot = 50;
            break;
        case 11: // short straight
            for(i = 0; i <= 6; i++)
                D[i] = 0;
            for(i = 0; i < 5; i++)
                D[dice[i]] = 1;
            for(i = 1; i <= 3; i++) {
                if(D[i] && D[i+1] && D[i+2] && D[i+3])
                    tot = 25;
            }
            break;
        case 12: // long straight
            for(i = 0; i < 4; i++) {
                if(dice[i] != dice[i+1]-1)
                    return 0;
            }
            tot = 35;
            break;
        case 13: //full house
            if(dice[0] == dice[1] && dice[2] == dice[4])
                tot = 40;
            if(dice[0] == dice[2] && dice[3] == dice[4])
                tot = 40;
            break;
    }
    return tot;
}
int count_bit(int n) {
    static int i, j;
    for(i = 0, j = 0; i < 13; i++)
        j += (n>>i)&1;
    return j;
}
int dp[STATE][STATE2]; // = max score
int score[DSIZE][DSIZE]; // score[i][j] = DICE[i] cat j
int arg_dp[STATE][STATE2][2]; //[0] category, [1] pre(1-6) total score
void sol_dp() {
    int i, j, k, p, q;
    for(i = 0; i < 13; i++)
        for(j = 0; j < 13; j++)
            score[i][j] = score_cat(DICE[i], j+1);
    memset(dp, -1, sizeof(dp));
    dp[0][0] = 0;
    int s, nstate, a, b;
    for(k = 0; k < 13; k++) { // add DICE[k]
        for(i = 0; i < STATE; i++) {// i = STATE mark
            if(count_bit(i) == k) {
                for(j = 0; j < 13; j++) {//set category
                    if(!((i>>j)&1)) {
                        s = score[k][j];
                        nstate = i|(1<<j); // new state
                        a = j < 6 ? s : 0;
                        for(p = 0; p < STATE2; p++) {
                            if(dp[i][p] >= 0) {
                                b = (p+a < 63) ? p+a : 63;
                                if(dp[nstate][b] < dp[i][p]+s) {
                                    dp[nstate][b] = dp[i][p]+s;
                                    arg_dp[nstate][b][0] = j;
                                    arg_dp[nstate][b][1] = p;
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    int mx = 0, bouns = 0;
    for(i = 0; i < 63; i++)
        if(dp[STATE-1][i] > mx) {
            mx = dp[STATE-1][i];
            b = i;
        }
    if(dp[STATE-1][63] >= 0 && dp[STATE-1][63]+35 > mx) {
        mx = dp[STATE-1][63]+35;
        b = 63;
        bouns = 35;
    }
    // backtracking
    int category[13], state = STATE-1;
    int ans[13];
    for(i = 12; i >= 0; i--) {
        category[i] = arg_dp[state][b][0];
        b = arg_dp[state][b][1];
        state ^= 1<<category[i];
        ans[category[i]] = score[i][category[i]];
    }
    for(i = 0; i < 13; i++)
        printf("%d ", ans[i]);
    printf("%d %d\n", bouns, mx);
}
int main() {
    int i, j;
    while(1) {
        for(i = 0; i < 13; i++) {
            for(j = 0; j < 5; j++) {
                if(scanf("%d", &DICE[i][j]) != 1)
                    return 0;
            }
            sort(DICE[i], DICE[i]+5);
        }
        //puts("==== START");
        sol_dp();
    }
    return 0;
}