【状压dp】AC Challenge

https://nanti.jisuanke.com/t/30994

把每道题的前置条件用二进制压缩，然后dp枚举所有可能状态，再枚举该状态是从哪一个节点转移来的，符合前置条件则更新。

代码：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1 << 20;
ll dp[maxn], a[21], b[21];
int pre[21];
int calc(int s) {
	int res = 0;
	for (int i = 0; i < 20; i++) {
		if ((1 << i) & s) {
			res++;
		}
	}
	return res;
}
int main() {
	int n;
	memset(dp, -0x3f, sizeof dp);
	scanf("%d", &n);
	for (int i = 1, t, x; i <= n; i++) {
		scanf("%lld%lld%d", &a[i], &b[i], &t);
		while (t--) {
			scanf("%d", &x);
			pre[i] += (1 << (x - 1));
		}
	}
	ll ans = 0;
	dp[0] = 0;
	for (int i = 1; i < (1 << n); i++) {
		for (int j = 0; j < n; j++) {
			if (i & (1 << j)) {
				int t = i - (1 << j);
				if ((t & pre[j + 1]) == pre[j + 1]) {
					dp[i] = max(dp[i], dp[t] + calc(i) * a[j + 1] + b[j + 1]);
					ans = max(ans, dp[i]);
				}
			}
		}
	}
	printf("%lld
", ans);
}