Happy Matt Friends HDU - 5119 （dp+滚动数组优化）

最新推荐文章于 2020-04-26 11:06:33 发布

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本文介绍了一个涉及游戏策略的问题，即如何从一组特定数值中选择若干个进行异或运算，使得结果不低于设定阈值，进而计算出所有可能的获胜策略数量。文章通过动态规划方法解决该问题，并提供了一段实现代码。

Matt has N friends. They are playing a game together.

Each of Matt’s friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends’magic numbers is no less than M , Matt wins.

Matt wants to know the number of ways to win.
Input
The first line contains only one integer T , which indicates the number of test cases.

For each test case, the first line contains two integers N, M (1 ≤ N ≤ 40, 0 ≤ M ≤ 10^6).

In the second line, there are N integers ki (0 ≤ k i ≤ 10^6), indicating the i-th friend’s magic number.
Output
For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y indicates the number of ways where Matt can win.
Sample Input
2
3 2
1 2 3
3 3
1 2 3
Sample Output
Case #1: 4
Case #2: 2

Hint
In the ﬁrst sample, Matt can win by selecting:
friend with number 1 and friend with number 2. The xor sum is 3.
friend with number 1 and friend with number 3. The xor sum is 2.
friend with number 2. The xor sum is 2.
friend with number 3. The xor sum is 3. Hence, the answer is 4.

大致题意：有n个数，然后然你从中选一些出来将他们全部异或起来，所得到的值如果不小于m 则合法，问有多少种合法的选择方案，你可以一个都不选。

思路：假设dp[i][j]表示从前i个数中选择一些数异或起来结果为j的方案数。对于当前位置i上的数，我们有两种选择，选或者不选，所以状态转移方程为 dp[i][j]=dp[i-1][j^a[i]]+dp[i-1][j]。
然后这里我们还需用滚动数组来优化内存。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<string>
#define LL long long  
using namespace std;
LL dp[2][1<<21];
int a[45];
int main()
{
    int T;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
        if(m==0)//如果为0，那么总的合法方案数即2^n
        {

            printf("Case #%d: %lld\n",cas,1<<n);
            continue;
        }
        memset(dp,0,sizeof(dp));

        dp[0][0]=1;
        for(int i=1;i<=n;i++)
        for(int j=0;j<(1<<21);j++)
        {
            dp[i&1][j]=dp[(i-1)&1][j]+dp[(i-1)&1][j^a[i]];
        }
        LL sum=0;
        for(int i=m;i<(1<<21);i++)
        sum+=dp[n&1][i];
        printf("Case #%d: %lld\n",cas,sum);
    }
    return 0;
}