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Problem solution · Go

Codeforces 2143D2 — Inversion Graph Coloring (Hard Version)

Codeforces 2143D2 — Inversion Graph Coloring (Hard Version): 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
66 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 2143D2 — Inversion Graph Coloring (Hard Version), 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

  • 66 lines of Go from the credited upstream file 2143D2.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks 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 2143D2 — Inversion Graph Coloring (Hard Version) · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io") // https://github.com/EndlessChengtype fenwick43 []int func (t fenwick43) update(i, v int) {	for ; i < len(t); i += i & -i {		t[i] += v	}} func (t fenwick43) pre(i int) (res int) {	for ; i > 0; i &= i - 1 {		res += t[i]	}	return} func cf2143D2(in io.Reader, out io.Writer) {	const mod = 1_000_000_007	var T, n, v int	for Fscan(in, &T); T > 0; T-- {		Fscan(in, &n)		row := make([]fenwick43, n+1)		col := make([]fenwick43, n+1)		for i := range col {			row[i] = make(fenwick43, n+1)			col[i] = make(fenwick43, n+1)		}		upd := func(x, y, v int) {			row[x].update(y, v)			col[y].update(x, v)		} 		// 定义 f[x][y] 表示表示第一个递增子序列以 x 结尾、第二个递增子序列以 y 结尾时，好子序列的数目		// 为避免重复统计，规定 x >= y		upd(1, 1, 1)		for range n {			Fscan(in, &v)			// 把 v 添加到第一个子序列的后面，必须满足 x <= v 且 y <= v			// 枚举 y，f[v][y] += sum_{x <= v} f[x][y]			for y := 1; y <= v; y++ {				upd(v, y, col[y].pre(v)%mod)			}			// 把 v 添加到第二个子序列的后面，必须满足 y <= v < x，这里 v != x 从而避免重复统计			// 枚举 x，f[x][v] += sum_{y <= v} f[x][y]			for x := v + 1; x <= n; x++ {				upd(x, v, row[x].pre(v)%mod)			}		} 		ans := 0		for _, f := range row {			ans += f.pre(n)		}		Fprintln(out, ans%mod)	}} //func main() { cf2143D2(bufio.NewReader(os.Stdin), os.Stdout) }

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