题目链接:https://www.luogu.org/problemnew/show/P1850
难的不在状态上,难在转移方程。
(话说方程写错居然还有84分= =)
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
using namespace std;
const ll maxn = 2010;
ll n, m, v, e, c[maxn], d[maxn], u[maxn][maxn], a, b, w;
double k[maxn], l[maxn], dp[maxn][maxn][2], ans = 1e9;
int main()
{
//freopen("change.in","r",stdin);
ios::sync_with_stdio(false);
memset(u, 63, sizeof(u));
memset(dp, 127, sizeof(dp));
cin>>n>>m>>v>>e;
for(ll i = 1; i <= n; i++)
cin>>c[i];
for(ll i = 1; i <= n; i++)
cin>>d[i];
for(ll i = 1; i <= n; i++)
{cin>>k[i]; l[i] = 1-k[i];}
for(ll i = 1; i <= e; i++)
{
cin>>a>>b>>w;
u[a][b] = min(w, u[a][b]);
u[b][a] = min(w, u[b][a]);
}
for(ll q = 1; q <= v; q++)
for(ll i = 1; i <= v; i++)
for(ll j = 1; j <= v; j++)
if(i != q && j != q && i != j) u[i][j] = min(u[i][j], u[i][q] + u[q][j]);
for(ll i = 1; i <= v; i++) u[i][i] = u[i][0] = u[0][i] = 0;
dp[1][0][0] = dp[1][1][1] = 0;
for(ll i = 2; i <= n; i++)
{
ll x1 = c[i], x2 = c[i-1], y1 = d[i], y2 = d[i-1];
dp[i][0][0] = dp[i-1][0][0] + u[x2][x1];
for(ll j = 0; j <= min(i, m); j++)
{
dp[i][j][0] = min(dp[i][j][0], min(dp[i-1][j][0] + u[x2][x1], dp[i-1][j][1] + l[i-1] * u[x2][x1] + k[i-1] * u[y2][x1]));
dp[i][j][1] = min(dp[i][j][1], min(dp[i-1][j-1][0] + u[x2][x1] * l[i] + u[x2][y1] * k[i], dp[i-1][j-1][1] + k[i-1] * l[i] * u[y2][x1] + l[i] * l[i-1] * u[x2][x1] + k[i] * k[i-1] * u[y2][y1] + l[i-1] * k[i] * u[x2][y1]));
}
}
for(ll i = 0; i <= m; i++)
ans = min(ans, min(dp[n][i][0], dp[n][i][1]));
printf("%0.2lf",ans);
return 0;
}