【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\n", ans);
}

【状压dp】AC Challenge

原文：https://www.cnblogs.com/stolf/p/9572209.html