思路
先floyed出两点最短路
然后就可以直接\(dp_{i,j,0/1}\)表示前i节课选择换j节，换不换当前这一节的最小贡献
直接可以枚举上一次决策的状态计算概率进行统计就可以了
我变量名写重了僵硬了半天。。。。被安排了
//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
//typename
typedef long long ll;
//convenient for
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
//inf of different typename
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
//fast read and write
template <typename T>
void Read(T &x) {
bool w = 1;x = 0;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-') w = 0, c = getchar();
while (isdigit(c)) {
x = (x<<1) + (x<<3) + c -'0';
c = getchar();
}
if (!w) x = -x;
}
template <typename T>
void Write(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) Write(x / 10);
putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = 2010;
const int M = 310;
double g[M][M], p[N];
double dp[N][N][2];
int c[N], d[N];
int n, m, v, e;
int main() {
Read(n), Read(m), Read(v), Read(e);
fu(i, 1, n) Read(c[i]);
fu(i, 1, n) Read(d[i]);
fu(i, 1, n) scanf("%lf", &p[i]);
fu(i, 1, v)
fu(j, 1, v) g[i][j] = 1e8;
fu(i, 1, e) {
int x, y; double w;
Read(x), Read(y); scanf("%lf", &w);
g[x][y] = g[y][x] = min(w, g[x][y]);
}
fu(i, 1, v) g[i][i] = 0.0;
fu(k, 1, v)
fu(i, 1, v) if (i != k)
fu(j, 1, v) if (j != k)
g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
double ans = 1e8;
fu(i, 1, n)
fu(j, 0, m) dp[i][j][0] = dp[i][j][1] = 1e8;
dp[1][0][0] = dp[1][1][1] = 0;
fu(i, 2, n) {
fu(j, 0, m) {
dp[i][j][0] = min(dp[i - 1][j][1] + p[i - 1] * g[d[i - 1]][c[i]] + (1.0 - p[i - 1]) * g[c[i - 1]][c[i]], dp[i - 1][j][0] + g[c[i - 1]][c[i]]);
if (j) dp[i][j][1] = min(dp[i - 1][j - 1][1] + p[i - 1] * p[i] * g[d[i - 1]][d[i]] + (1.0 - p[i - 1]) * p[i] * g[c[i - 1]][d[i]] + p[i - 1] * (1.0 - p[i]) * g[d[i - 1]][c[i]] + (1.0 - p[i - 1]) * (1.0 - p[i]) * g[c[i - 1]][c[i]], dp[i - 1][j - 1][0] + (p[i] * g[c[i - 1]][d[i]] + (1.0 - p[i]) * g[c[i - 1]][c[i]]));
}
}
fu(i, 0, m) ans = min(ans, min(dp[n][i][0], dp[n][i][1]));
printf("%.2lf", ans);
return 0;
}