codeforces 580D . Kefa and Dishes 状压dp

题目链接

 

令dp[i][j]表示状态i中最后一个拿的是j的情况， 然后转移就好。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int val[20], m, g[20][20];
ll dp[1<<18][18];
int main()
{
    int n, k, x, y, z;
    cin>>n>>m>>k;
    for(int i = 0; i<n; i++) {
        scanf("%d", &val[i]);
    }
    for(int i = 0; i<k; i++) {
        scanf("%d%d%d", &x, &y, &z);
        g[x-1][y-1] = z;
    }
    mem1(dp);
    for(int i = 0; i<n; i++) {
        dp[1<<i][i] = val[i];
    }
    ll ans = 0;
    for(int i = 0; i<(1<<n); i++) {
        int cnt = 0, flag = 0;
        for(int j = 0; j<n; j++)
            if(1<<j&i)
                cnt++;
        if(cnt == m)
            flag = 1;
        for(int j = 0; j<n; j++) {
            if(dp[i][j]==-1)
                continue;
            if(flag)
                ans = max(ans, dp[i][j]);
            for(int k = 0; k<n; k++) {
                if(1<<k&i)
                    continue;
                dp[i|1<<k][k] = max(dp[i][j]+val[k]+g[j][k], dp[i|1<<k][k]);
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}