[Dp每日训练]送外卖

题目：送外卖2

分析：

每个任务有三个状态，未开始，已开始未结束，已结束。考虑状压。
dp[i][j]表示所有任务的状态，将每个任务用三进制表示成i，j表示此时在j点，所花费的最少时间。
状态转移的时候先枚举每个任务，然后枚举转移过来的点。

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;//三年竞赛一场空，不开long long见祖宗 
//typedef __int128 lll;
#define print(i) cout << "debug: " << i << endsl
#define close() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define mem(a, b) memset(a, b, sizeof(a))
#define pb(a) push_back(a)
#define x first
#define y second
typedef pair<int, int> pii;
const ll mod = 1e9 + 7;
const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const int N = 30;
struct node
{
    int u, v, start, end;
}a[N];
int power[N];
int dp[60000][N];
int dis[N][N];
int n, m, q; 

void init()
{
    power[0] = 1;
    for(int i = 1; i < N; i++)
        power[i] = power[i - 1] * 3;
}

void floyd()
{
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
}

int main()
{
    init();
    cin >> n >> m >> q;
    mem(dis, inf);
    for(int i = 1; i <= n; i++) dis[i][i] = 0;
    for(int i = 1; i <= m; i++)
    {
        int a, b, val; cin >> a >> b >> val;
        dis[a][b] = min(dis[a][b], val);
    }
    floyd();
    for(int i = 0; i < q; i++)
        cin >> a[i].u >> a[i].v >> a[i].start >> a[i].end;

    mem(dp, inf); dp[0][1] = 0;
    for(int i = 1; i < power[q]; i++)
        for(int j = 0; j < q; j++)
            for(int k = 1; k <= n; k++)
            {
                if(i % power[j + 1] / power[j] == 1)
                    dp[i][a[j].u] = min(dp[i][a[j].u], max(dp[i - power[j]][k] + dis[k][a[j].u], a[j].start));
                else if(i % power[j + 1] / power[j] == 2 && dp[i - power[j]][k] + dis[k][a[j].v] <= a[j].end)
                    dp[i][a[j].v] = min(dp[i][a[j].v], dp[i - power[j]][k] + dis[k][a[j].v]);
            }
    int res = 0;
    for(int i = 0; i < power[q]; i++)
        for(int j = 1; j <= n; j++)
        {
            if(dp[i][j] == inf) continue;
            int num = 0;
            for(int k = 0; k < q; k++)
                if(i % power[k + 1] / power[k] == 2)
                    num++;
            res = max(res, num);
        }
    cout << res << endl;
}