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UVA 1558 - Number Game

题目链接

题意：20之内的数字，每次可以选一个数字，然后它的倍数，还有其他已选数的倍数组合的数都不能再选，谁先不能选数谁就输了，问赢的方法

思路：利用dp记忆化去求解，要输出方案就枚举第一步即可，状态转移过程中，选中一个数字，相应的变化写成一个函数，然后就是普通的博弈问题了，必胜态之后必有必败态，必败态之后全是必胜态

代码：

#include <stdio.h>
#include <string.h>

const int N = 1050005;
int t, n, w, start, dp[N], ans[25], an;

int getnext(int state, int x) {
	for (int i = x; i <= 20; i += x) 
 		if (state&(1<<(i - 2)))
   			state ^= (1<<(i - 2));
	for (int i = 2; i <= 20; i++) {
		if (state&(1<<(i - 2))) {
			for (int j = x; i - j >= 2; j += x) {
				if (!(state&(1<<(i - j - 2)))) {
					state ^= (1<<(i - 2)); 
					break;
				}
   			}
  		}
 	}
 	return state;
}

int dfs(int state) {
	if (dp[state] != -1) return dp[state];
	if (state == 0) return dp[state] = 0;
	
	for (int i = 2; i <= 20; i++) {
		if (state&(1<<(i - 2))) {
			if (dfs(getnext(state, i)) == 0)
				return dp[state] = 1;
  		}
 	}
 	return dp[state] = 0;
}

int main() {
	int cas = 0;
	scanf("%d", &t);
	memset(dp, -1, sizeof(dp));
	while (t--) {
		start = 0; an = 0;
		scanf("%d", &n);
		for (int i = 0; i < n; i++) {
			scanf("%d", &w);
			start |= (1<<(w - 2));
  		}
  		for (int i = 2; i <= 20; i++) {
  			if (start&(1<<(i - 2))) {
  				if (dfs(getnext(start, i)) == 0)
  					ans[an++] = i;
     		}
    	}
    	printf("Scenario #%d:\n", ++cas);
    	if (an) {
    		printf("The winning moves are:");
    		for (int i = 0; i < an; i++)
    			printf(" %d", ans[i]);
  			printf(".\n");
     	}
     	else printf("There is no winning move.\n");
     	printf("\n");
 	}
	return 0;
}

UVA 1558 - Number Game(博弈dp),布布扣,bubuko.com

UVA 1558 - Number Game(博弈dp)

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原文地址：http://blog.csdn.net/accelerator_/article/details/37889953