2018icpc南京网络赛-E AC Challenge(状压+dfs)

题意：

n道题，每道题有ai和bi，完成这道题需要先完成若干道题，完成这道题可以得到分数t*ai+bi，其中t是时间

1s, n<=20

思路：

由n的范围状压，状态最多1e6

然后dfs，注意代码中dfs里的剪枝，

对一个状态statu，因为贪心的取最大值就行，所以及时剪枝

代码：

当时写不出来真是菜的活该

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const ll Inf = 0x3f3f3f3f3f3f3f3f;
const db pi = acos(-1.0);
int n;
ll dp[maxn];//达到status的最大值
ll a[30], b[30];
int s[30];
vector<int>nd[30];
void init(){
    for(int i = 1; i <= n; i++){
        int ans = 0;
        int sz = nd[i].size();
        for(int j = 0; j < sz; j++){
            ans|=(1<<(nd[i][j]-1));
        }
        s[i]=ans;
    }
    return;
}
void dfs(int cnt, int statu, ll ans){
    dp[statu] = max(dp[statu], ans);
    if(cnt == n)return;
    for(int i = 1; i <= n; i++){
        if(((1<<(i-1))&statu)!=0)continue;
        if((s[i]&statu)!=s[i])continue;
        dfs(cnt+1,statu|(1<<(i-1)), ans + (cnt+1)*a[i]+b[i]);
    }
    return;
}
int main(){
    mem(dp,-1);
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        int num;
        scanf("%lld %lld %d", &a[i], &b[i], &num);
        for(int j = 0; j < num; j++){
            int x;
            scanf("%d",&x);
            nd[i].pb(x);
        }
    }
    init();
    dfs(0,0,0);
    printf("%lld",dp[(1<<n)-1]<0?0:dp[(1<<n)-1]);
    return 0;
}

/*

 */

原文地址：https://www.cnblogs.com/wrjlinkkkkkk/p/10182537.html