中山网站制作专业网络广告营销的典型案例

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;
}