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d.go
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d.go
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package main
// github.com/EndlessCheng/codeforces-go
func minCost(maxTime int, edges [][]int, fees []int) int {
const inf int = 1e9
n := len(fees)
dp := make([][]int, maxTime+1)
for i := range dp {
dp[i] = make([]int, n)
for j := range dp[i] {
dp[i][j] = inf
}
}
dp[0][0] = fees[0]
// 原图是存在环的,如果直接在拆点图上跑最短路是需要用 Dijkstra 等最短路算法的
// 但是注意到,若按照时间的升序转移,由于图中边权均为正,从当前时间出发是不可能转移到过去的时间上的,从而保证状态无后效性,也就无需使用最短路算法来求解了
for t := 1; t <= maxTime; t++ {
for _, e := range edges {
if e[2] <= t {
v, w, wt := e[0], e[1], e[2]
dp[t][v] = min(dp[t][v], dp[t-wt][w]+fees[v])
dp[t][w] = min(dp[t][w], dp[t-wt][v]+fees[w])
}
}
}
ans := inf
for _, dv := range dp[1:] {
ans = min(ans, dv[n-1])
}
if ans < inf {
return ans
}
return -1
}
func min(a, b int) int {
if a < b {
return a
}
return b
}