codeforces 580d 状压DP

题意：有n种菜，现在选m种菜来吃，如果在吃y的前一道菜是x的话，那么就可以获得额外满意度。每一种菜都有一个满意度。

思路：设dp[i][S]表示为最后一道菜为i，现在的菜吃的状态为S。S中的二进位如果为1表示已经吃了，如果是0则表示没吃，状压DP，答案就出了。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
using namespace std;
#define EPS 1e-10
typedef long long ll;
const int maxn = 1<<20;
const int INF = 1e9 ;
const double eps = 1e-8;
const int mod = 2520;
int gra[20][20],a[20],num[20],cnt;
ll dp[maxn][20];
int cal(int u){
    cnt = 0;
    int cntp = 0;
    while(u > 0){
        if(u % 2 == 1) num[++cnt] = cntp;
        cntp++;
        u /= 2;
    }
    return cnt;
}
int main(){
  //  freopen("in.txt","r",stdin);
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    for(int i = 1;i <= n;i++)
        scanf("%d",&a[i]);
    int se = (1<<n) - 1;
    int a1,b,c;
    for(int i = 1;i <= k;i++){
        scanf("%d%d%d",&a1,&b,&c);
        gra[a1][b] = c;
    }
    ll smax = 0;

    for(int i = 1;i <= se;i++){
        if(cal(i) > m) continue;
        ll sum = 0;
        for(int j = 1;j <= cnt;j++){
            int posj = num[j];
           // cout<<posj+1<<" ";
           // int posj = num[j];
           // dp[i][posj] = sum;
            for(int k = 1;k <= cnt;k++){
                if(k == j&&cnt != 1) continue;
                int posk = num[k];
                dp[i][posj] = max(dp[i][posj],dp[i-(1<<posj)][posk] + gra[posk+1][posj+1] + a[posj+1]);
               // cout<<posk+1<<" "<<dp[i][posj]<<" "<<"b";
            }

           // cout<<endl;
            smax = max(smax,dp[i][posj]);
        }
    }
    printf("%I64d
",smax);
    return 0;
}