【luogu2583】地铁间谍

https://www.luogu.org/problem/show?pid=2583

设在时刻i，车站j到结束最短需要等待f(i,j)分钟，得状态转移方程：

f(i,j)=min{
     f(i+1,j)+1
     f(i+t[j-1],j-1) (若时刻i有向左走的车)
     f(i+t[j],j+1) (若时刻i有向右走的车)
}
因此需要预处理每一时刻有没有车可以坐。

初始值f(t,n)=0，其他均为∞。

#include <iostream>
#include <cstring>
#define maxt 210
#define maxn 60
#define inf 1000
using namespace std;
bool has_train[maxt][maxn][2];
int dp[maxt][maxn];
int n, tt, t[maxn], m1, m2, d;
int main()
{
    int cnt = 1;
    while (true)
    {
        cin >> n;
        if (n == 0) break;
        cin >> tt;
        for (int i = 1; i<n; i++) cin >> t[i];

        //预处理has_train数组
        memset(has_train, false, maxt * maxn * 2);
        cin >> m1;
        while (m1--)
        {
            cin >> d;
            for (int j = 1; j <= n - 1 && d <= tt; j++)
            {
                has_train[d][j][0] = true;
                d += t[j];
            }
        }
        cin >> m2;
        while (m2--)
        {
            cin >> d;
            for (int j = n; j >= 2 && d <= tt; j--)
            {
                has_train[d][j][1] = true;
                d += t[j - 1];
            }
        }

        //初始化dp数组
        for (int i = 1; i <= n; i++)
            dp[tt][i] = inf;
        dp[tt][n] = 0;

        for (int i = tt - 1; i >= 0; i--)
        {
            for (int j = 1; j <= n; j++)
            {
                dp[i][j] = dp[i + 1][j] + 1; //+1s
                if (i + t[j] <= tt && j + 1 <= n && has_train[i][j][0])
                    dp[i][j] = min(dp[i][j], dp[i + t[j]][j + 1]); //向右走
                if (i + t[j - 1] <= tt && j - 1 >= 1 && has_train[i][j][1])
                    dp[i][j] = min(dp[i][j], dp[i + t[j - 1]][j - 1]); //向左走
            }
        }
        if (dp[0][1]<inf)
            cout << "Case Number " << cnt << ": " << dp[0][1] << endl;
        else
            cout << "Case Number " << cnt << ": impossible" << endl;
        cnt++;
    }
    return 0;
}