ACM-ICPC 2018 南京赛区网络预赛 - AC Challenge（状压DP）

ACM-ICPC 2018 南京赛区网络预赛 - AC Challenge

题意：

有n个题目，每个题目有一些信息， $a_i ,b_i, s_i, [p_{i,1},p_{i,2},p_{i,3},..,p_{i,s}]$ ，第 t 个过第 i 题会得到分数 t*ai + bi

在过第 i 题前必须要先过 $p_{i,1},p_{i,2},p_{i,3},..,p_{i,s}$ 这几题，可以选择不过题

求最大得分

$0 < n \leq 20, 0<p_{i,j} \leq n,0< j \leq s_i , 0 < i \leq n , |ai| \leq 10^9, |bi| \leq 10^9$

最多20个题目，不计顺序的话，一共有 2^20 = 1,048,576 种取法

做出 i 个可以从做出 i-1 个的状态推出，枚举第 i 个做出的题目，得到每个做出 i 个题目的最大值然后继续下推

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
typedef long long int LL;
LL n, x, s, ans, dp[1<<21];
struct node{LL a, b, p;}a[22];
LL count(LL x, int t) {LL ans = -1e12;for(int i=0;i<n;i++) {if(x & (1<<i)) {if((x & a[i+1].p) == a[i+1].p)ans = max(ans, dp[x ^ (1<<i)] + t*a[i+1].a + a[i+1].b);}}return dp[x] = ans;
}
void f(int x, int pre, int cnt, int tot) {if(cnt == x) {ans = max(ans, count(tot, x));return;}if(n-pre < x-cnt) return;for(int i=pre+1;i<=n;i++) f(x, i, cnt+1, tot | (1<<(i-1)));
}
int main()
{scanf("%lld", &n);for(int i=1;i<=n;i++) {scanf("%lld%lld%lld", &a[i].a, &a[i].b, &s);for(int j=1;j<=s;j++) {scanf("%lld", &x); a[i].p |= (1<<(x-1));}}for(int i=1;i<=n;i++) f(i, 0, 0, 0);printf("%lld\n", ans);return 0;
}

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