Solving 10385 - Duathlon (Ternary search)
[and.git] / 10149 - Yahtzee / 10149.2.cpp
blob9b3e51cb465a163629147b344262ea2e9d55e12a
1 using namespace std;
2 #include <algorithm>
3 #include <iostream>
4 #include <iterator>
5 #include <sstream>
6 #include <fstream>
7 #include <numeric>
8 #include <cassert>
9 #include <climits>
10 #include <cstdlib>
11 #include <cstring>
12 #include <string>
13 #include <cstdio>
14 #include <vector>
15 #include <cmath>
16 #include <queue>
17 #include <deque>
18 #include <stack>
19 #include <list>
20 #include <map>
21 #include <set>
23 #define foreach(x, v) for (typeof (v).begin() x = (v).begin(); x != (v).end(); ++x)
24 #define For(i, a, b) for (int i=(a); i<(b); ++i)
25 #define D(x) cout << #x " is " << x << endl
27 typedef int turn[5];
29 enum category {
30 ones, twos, threes, fours, fives, sixes, chance, three_of_a_kind,
31 four_of_a_kind, five_of_a_kind, short_straight, long_straight,
32 full_house
35 int score(turn t, int c){
36 int times[7] = {0};
37 for (int i=0; i<5; ++i) times[t[i]]++;
38 switch(c){
39 case ones:
40 case twos:
41 case threes:
42 case fours:
43 case fives:
44 case sixes:
45 return times[c+1] * (c+1);
46 case chance:
47 return accumulate(t, t+5, 0);
48 case three_of_a_kind:
49 for (int i=1; i<=6; ++i) if (times[i] >= 3) return accumulate(t, t+5, 0);
50 return 0;
51 case four_of_a_kind:
52 for (int i=1; i<=6; ++i) if (times[i] >= 4) return accumulate(t, t+5, 0);
53 return 0;
54 case five_of_a_kind:
55 for (int i=1; i<=6; ++i) if (times[i] >= 5) return 50;
56 return 0;
57 case short_straight:
58 for (int i=1; i<=3; ++i)
59 if (times[i] > 0 && times[i+1] > 0 && times[i+2] > 0 && times[i+3] > 0) return 25;
60 return 0;
61 case long_straight:
62 for (int i=1; i<=2; ++i)
63 if (times[i] > 0 && times[i+1] > 0 && times[i+2] > 0 && times[i+3] > 0 && times[i+4] > 0) return 35;
64 return 0;
66 case full_house:
67 for (int i=1; i<=6; ++i)
68 for (int j=1; j<=6; ++j)
69 if (times[i] == 3 && times[j] == 2) return 40;
70 return 0;
71 default:
72 printf("Something very bad happened.\n");
74 return 0;
77 turn turns[13];
79 int dp[13][1 << 13];
81 dp[i][mask] = Best score having matched categories [0..i] and turns <on> on mask.
83 int choice[13][1 << 13];
85 int main(){
86 while (true){
87 for (int i=0; i<13; ++i){
88 for (int j=0; j<5; ++j){
89 if (scanf("%d", &turns[i][j]) != 1) return 0;
93 for (int i=0; i<13; ++i)
94 for (int j=0; j<(1 << 13); ++j)
95 dp[i][j] = INT_MIN, choice[i][j] = -1;
98 //First category
99 for (int t=0; t<13; ++t){
100 dp[ones][1 << t] = score(turns[t], ones);
101 choice[ones][1 << t] = t;
103 //Let's dance!
104 for (int c=twos; c<=full_house; ++c){
105 for (int mask = 0; mask < (1 << 13); ++mask){
106 if (__builtin_popcount(mask) != c + 1) continue;
107 for (int t=0; t<13; ++t){
108 int prev_mask = mask & ~(1 << t);
110 int canDo = dp[c-1][prev_mask] + score(turns[t], c);
111 if (canDo > dp[c][mask]){
112 dp[c][mask] = canDo;
113 choice[c][mask] = t;
116 if (c == sixes && dp[c][mask] >= 63){
117 //Add bonus
118 dp[c][mask] += 35;
123 //Now find the matching from the computed table
124 //The last state is dp[all_categories][all_turns] = dp[12][11111...1111B]
125 int match[13];
126 for (int c=12, mask = (1 << 13) - 1; c >= 0; c--){
127 match[c] = choice[c][mask];
128 mask &= ~(1 << choice[c][mask]);
130 int sum = 0, bonus = 0;
131 for (int c = ones; c <= full_house; ++c){
132 int s = score(turns[match[c]], c);
133 printf("%d ", s);
134 sum += s;
135 if (c == sixes && sum >= 63){
136 sum += (bonus = 35);
140 //assert(sum == dp[12][(1 << 13) - 1]);
141 printf("%d %d\n", bonus, sum);
143 return 0;