POJ 1143 Number Game(状压DP)

Description

Christine and Matt are playing an exciting game they just invented: the Number Game. The rules of this game are as follows. 
The players take turns choosing integers greater than 1. First, Christine chooses a number, then Matt chooses a number, then Christine again, and so on. The following rules restrict how new numbers may be chosen by the two players: 

• A number which has already been selected by Christine or Matt, or a multiple of such a number,cannot be chosen. 
  
• A sum of such multiples cannot be chosen, either.

If a player cannot choose any new number according to these rules, then that player loses the game. 
Here is an example: Christine starts by choosing 4. This prevents Matt from choosing 4, 8, 12, etc.Let's assume that his move is 3. Now the numbers 3, 6, 9, etc. are excluded, too; furthermore, numbers like: 7 = 3+4;10 = 2*3+4;11 = 3+2*4;13 = 3*3+4;... are also not available. So, in fact, the only numbers left are 2 and 5. Christine now selects 2. Since 5=2+3 is now forbidden, she wins because there is no number left for Matt to choose. 
Your task is to write a program which will help play (and win!) the Number Game. Of course, there might be an infinite number of choices for a player, so it may not be easy to find the best move among these possibilities. But after playing for some time, the number of remaining choices becomes finite, and that is the point where your program can help. Given a game position (a list of numbers which are not yet forbidden), your program should output all winning moves. 
A winning move is a move by which the player who is about to move can force a win, no matter what the other player will do afterwards. More formally, a winning move can be defined as follows. 

• A winning move is a move after which the game position is a losing position. 
  
• A winning position is a position in which a winning move exists. A losing position is a position in which no winning move exists. 
  
• In particular, the position in which all numbers are forbidden is a losing position. (This makes sense since the player who would have to move in that case loses the game.)

Input

The input consists of several test cases. Each test case is given by exactly one line describing one position. 
Each line will start with a number n (1 <= n <= 20), the number of integers which are still available. The remainder of this line contains the list of these numbers a1;...;an(2 <= ai <= 20). 
The positions described in this way will always be positions which can really occur in the actual Number Game. For example, if 3 is not in the list of allowed numbers, 6 is not in the list, either. 
At the end of the input, there will be a line containing only a zero (instead of n); this line should not be processed.

Output

For each test case, your program should output "Test case #m", where m is the number of the test case (starting with 1). Follow this by either "There's no winning move." if this is true for the position described in the input file, or "The winning moves are: w1 w2 ... wk" where the wi are all winning moves in this position, satisfying wi < wi+1 for 1 <= i < k. After this line, output a blank line.

Sample Input

2 2 52 2 35 2 3 4 5 60

Sample Output

Test Case #1The winning moves are: 2Test Case #2There's no winning move.Test Case #3The winning moves are: 4 5 6

分析：

这是一个博弈问题，只需要从终止状态开始，将每一个可能的状态标记为必胜态或必败态。终止状态为必败态，可以一步转移到必败态的是必胜态，否则是必败态。因为所有数不超过20，可以采用状态压缩表示每一个状态：第i位为1表示i在该状态下可以取，0表示不可以取。注意在考虑状态转移的时候需要判断状态是否合法：若x可以取，则对于x=y+z，其中y,z至少为2，y,z之一也是可以取的。

代码如下。

/*PROG: POJ1143PROB: DP + state compression*/#include <cstdio>#include <vector>#include <cstring>#include <algorithm>using namespace std;#define DEBUG 1#define LOG(...) do { if (DEBUG) fprintf(stderr, __VA_ARGS__); } while(0)#define MAXN 1<<22char c[23], d[23];char dp[MAXN];int main(void) {int kase = 0;int n;while (scanf("%d", &n), n) {printf("Test Case #%d\n", ++kase);vector<int> a(n);for (int i = 0; i < n; ++i) {int tmp; scanf("%d", &tmp);a[i] = tmp;}sort(a.begin(), a.end());memset(dp, 0, sizeof(dp));memset(c, 0, sizeof(c));for (int i = 0; i < n; ++i)c[a[i]] = 1;int N = 1<<n;for (int i = 0; i < N; ++i) {memset(d, 0, sizeof(d));for (int j = 0; j < n; ++j)d[a[j]] = (i>>j)&1;// whether the state is validbool valid = true;for (int j = 0; j < n; ++j) {if (!d[a[j]]) continue;for (int k1 = 2, k2 = a[j]-2; k1 <= k2; ++k1, --k2)if (!d[k1]&&!d[k2]) {valid = false; break;}if (!valid) break;}if (!valid) continue;bool win = false;for (int j = 0; j < n; ++j) {if (!d[a[j]]) continue;int tmp = i;tmp &= ~(1<<j);for (int j1 = j+1; j1 < n; ++j1) {if (!d[a[j1]]) continue;bool change = false;for (int k = a[j1]-a[j]; k > 1; k -= a[j])if (!d[k]) {change = true; break;}if (a[j1]%a[j]==0) change = true;if (change) tmp &= ~(1<<j1);}if (!dp[tmp]) {win = true; break;}}dp[i] = win?1:0;// LOG("dp[%d]=%d\n", i, dp[i]);}int x = (1<<n)-1;vector<int> ok;for (int i = 0; i < n; ++i) {int tmp = x;tmp &= ~(1<<i);for (int j = i+1; j < n; ++j) {bool change = false;for (int k = a[j]-a[i]; k > 1; k -= a[i])if (!c[k]) {change = true; break;}if (a[j]%a[i]==0) change = true;if (change) tmp &= ~(1<<j);}// LOG("choose %d, tmp:%d\n", a[i], tmp);if (!dp[tmp]) ok.push_back(a[i]);}if (!ok.size()) printf("There's no winning move.\n\n");else {printf("The winning moves are:");for (int i = 0; i < ok.size(); ++i)printf(" %d", ok[i]);printf("\n\n");}}return 0;}