WOJ 1608 Calculation（状态压缩）

题意：给出n个数（n<=14），现在可以将这n个数任意分组，每组的每个数可以在前面加上一个正号或者符号，问最多有多少组的和等于s。
思路：状态压缩，每个状态表示当前的状态可以分成多少组含有s的组，然后状态转移即可。

#include<bits/stdc++.h>
#define eps 1e-6
#define LL long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

const int MAXN = 1<<15;
//const int INF = 0x3f3f3f3f;
int n, s, a[16], dp[MAXN], sumv[MAXN];
int main()
{
freopen("input.txt", "r", stdin);
int T;
scanf("%d", &T);
while (T--) {
scanf("%d%d", &n, &s);
memset(dp, 0, sizeof(dp));
memset(sumv, 0, sizeof(sumv));
for (int i = 0; i < n; i++)
scanf("%d", &a[i]);
for (int i = 0; i < (1<<n); i++)
for (int j = 0; j < n; j++)
if ((1<<j)&i)
sumv[i] += a[j];
for (int i = 0; i < (1<<n); i++) {
dp[i] = (sumv[i] == s);
for (int j = i; j; j = (j-1)&i)
dp[i] |= (sumv[i^j]-sumv[j] == s);
}
dp[0] = 0;
for (int i = 0; i < (1<<n); i++)
for (int j = i; j; j = (j-1)&i)
dp[i] = max(dp[i], dp[j]+dp[i^j]);
printf("%d\n", dp[(1<<n)-1]);
}
return 0;
}