CPMagical CP Library

Problem solution · Go

Codeforces 1245D — Shichikuji and Power Grid

Codeforces 1245D — Shichikuji and Power Grid: a Go solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Direct simulation
Source
EndlessCheng Codeforces Go
Length
104 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Codeforces 1245D — Shichikuji and Power Grid, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 104 lines of Go from the credited upstream file 1245D.go.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from EndlessCheng Codeforces Go by Σndless (EndlessCheng) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCodeforces 1245D — Shichikuji and Power Grid · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"math") // github.com/EndlessCheng/codeforces-gofunc CF1245D(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush()	abs := func(x int) int {		if x < 0 {			return -x		}		return x	} 	var n int	Fscan(in, &n)	type pair struct{ x, y int }	city := make([]pair, n)	for i := range city {		Fscan(in, &city[i].x, &city[i].y)	}	cityCost := make([]int, n)	for i := range cityCost {		Fscan(in, &cityCost[i])	}	wireCost := make([]int, n)	for i := range wireCost {		Fscan(in, &wireCost[i])	} 	// 构建距离（费用）矩阵	dis := make([][]int64, n+1)	for i := range dis {		dis[i] = make([]int64, n+1)	}	for i, c := range cityCost {		dis[i][n] = int64(c) // 建立虚点 n，将所有城市连到点 n 上，值为在该城市上建立发电站的费用		dis[n][i] = int64(c)		for j := i + 1; j < n; j++ {			// 搭建电线的费用			dis[i][j] = int64(abs(city[i].x-city[j].x)+abs(city[i].y-city[j].y)) * int64(wireCost[i]+wireCost[j])			dis[j][i] = dis[i][j]		}	} 	totCost := int64(0)	ansCity := []interface{}{}	ansWire := []pair{} 	// 由于是稠密图，用 Prim 算法求最小生成树	type pair2 struct {		cost int64		v    int	}	minCost := make([]pair2, n+1)	for i := range minCost {		minCost[i].cost = math.MaxInt64	}	minCost[0].cost = 0	used := make([]bool, n+1)	for {		v := -1		for w, b := range used {			if !b && (v < 0 || minCost[w].cost < minCost[v].cost) {				v = w			}		}		if v < 0 {			break		}		used[v] = true		totCost += minCost[v].cost		if v == n { // 发电			ansCity = append(ansCity, minCost[v].v+1)		} else if minCost[v].v == n { // 发电			ansCity = append(ansCity, v+1)		} else if minCost[v].v != v { // 连电线，注意避免自环			ansWire = append(ansWire, pair{minCost[v].v + 1, v + 1})		}		for w, d := range dis[v] {			if d < minCost[w].cost {				minCost[w] = pair2{d, v}			}		}	} 	Fprintln(out, totCost)	Fprintln(out, len(ansCity))	Fprintln(out, ansCity...)	Fprintln(out, len(ansWire))	for _, conn := range ansWire {		Fprintln(out, conn.x, conn.y)	}} //func main() { CF1245D(os.Stdin, os.Stdout) }

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗