洛谷 P1850 [NOIP 2016 提高组] 换教室（期望DP）【 提高+/省选−】

题目链接

https://www.luogu.com.cn/problem/P1850

思路

令$f[i][j][0/1]$分别表示已经上完前$i$节课，使用了$j$次交换申请，且第$i$节课不适用交换申请$or$第$i$节课使用交换申请的移动耗费的体力值的总和的期望值的最小值。

令$dist[i][j]$表示从第$i$个教室到第$j$个教室的最短路，因为数据范围较小，使用$floyd$求解即可。

则状态转移方程为：

$f[i][j][0]=min(f[i][j][0],f[i-1][j][0]+dist[c[i-1]][c[i]])$。

$f[i][j][0]=min(f[i][j][0],f[i-1][j][1]+(1-k[i-1])\ast dist[c[i-1]][c[i]]+k[i-1]\ast dist[d[i-1]][c[i]])$。

$f[i][j][1]=min(f[i][j][1],f[i-1][j-1][0]+(1-k[i])\ast dist[c[i-1]][c[i]]+k[i]\ast dist[c[i-1]][d[i]])$

$f[i][j][1]=min(f[i][j][1],f[i-1][j-1][1]+(1-k[i-1])\ast (1-k[i])\ast dist[c[i-1]][c[i]]+k[i-1]\ast (1-k[i])\ast dist[d[i-1]][c[i]]+(1-k[i-1])\ast k[i]\ast dist[c[i-1]][d[i]]+k[i-1]\ast k[i]\ast dist[d[i-1]][d[i]])$。

代码

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define double long double
#define endl "\n"

typedef long long i64;
typedef unsigned long long u64;
typedef pair<int, int> pii;

const int N = 2e3 + 5, M = 3e2 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9;
const double eps = 1e-6;

std::mt19937 rnd(time(0));

int n, m, v, e;
int c[N], d[N], dist[M][M];
double k[N];
double f[N][N][2];
void floyd(int maxx)
{
    for (int k = 1; k <= maxx; k++)
    {
        for (int i = 1; i <= maxx; i++)
        {
            for (int j = 1; j <= maxx; j++)
            {
                dist[i][j] = min(dist[i][k] + dist[k][j], dist[i][j]);
            }
        }
    }
}
void solve(int test_case)
{
    cin >> n >> m >> v >> e;
    for (int i = 1; i <= n; i++)
    {
        cin >> c[i];
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> d[i];
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> k[i];
    }
    int maxx = 300;
    for (int i = 1; i <= maxx; i++)
    {
        for (int j = 1; j <= maxx; j++)
        {
            dist[i][j] = inf;
        }
    }
    for (int i = 1; i <= maxx; i++)
        dist[i][i] = 0;
    for (int i = 1, a, b, w; i <= e; i++)
    {
        cin >> a >> b >> w;
        dist[a][b] = min(dist[a][b], w);
        dist[b][a] = min(dist[b][a], w);
    }
    floyd(maxx);
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= min(i, m); j++)
        {
            f[i][j][0] = f[i][j][1] = (double)(1e18);
        }
    }
    f[1][0][0] = 0;
    f[1][1][1] = 0;
    for (int i = 2; i <= n; i++)
    {
        for (int j = 0; j <= min(i, m); j++)
        {
            double res = 0;
            if (j < i)
            {
                f[i][j][0] = min(f[i][j][0], f[i - 1][j][0] + dist[c[i - 1]][c[i]]);
                res = f[i - 1][j][1] + ((double)(1) - k[i - 1]) * dist[c[i - 1]][c[i]] + k[i - 1] * dist[d[i - 1]][c[i]];
                f[i][j][0] = min(f[i][j][0], res);
            }

            if (j)
            {
                res = f[i - 1][j - 1][0] + ((double)(1) - k[i]) * dist[c[i - 1]][c[i]] + k[i] * dist[c[i - 1]][d[i]];
                f[i][j][1] = min(f[i][j][1], res);
                res = f[i - 1][j - 1][1];
                res += ((double)(1) - k[i - 1]) * ((double)(1) - k[i]) * dist[c[i - 1]][c[i]];
                res += k[i - 1] * ((double)(1) - k[i]) * dist[d[i - 1]][c[i]];
                res += ((double)(1) - k[i - 1]) * k[i] * dist[c[i - 1]][d[i]];
                res += k[i - 1] * k[i] * dist[d[i - 1]][d[i]];
                f[i][j][1] = min(f[i][j][1], res);
            }
        }
    }
    double ans = 1e18;
    for (int i = 0; i <= m; i++)
    {
        ans = min({ans, f[n][i][0], f[n][i][1]});
    }
    cout << fixed << setprecision(2) << ans << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test ; i++)
    {
        solve(i);
    }
    return 0;
}