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| 1 | +let input = '' |
| 2 | + |
| 3 | +process.stdin.on('data', data => input += data) |
| 4 | +process.stdin.on('end', () => { |
| 5 | + const lines = input.split('\n') |
| 6 | + const [t] = lines[0].split(' ').map(Number) |
| 7 | + for (let kase = 1; kase <= t; kase++) { |
| 8 | + const [_, d] = lines[kase * 2 - 1].split(' ').map(Number) |
| 9 | + const v = lines[kase * 2].split(' ').map(Number) |
| 10 | + const n = v.length |
| 11 | + /* |
| 12 | + dp[i][j][k] = 将区间 i, j 全部变成 k 的开销 |
| 13 | + Math.min(dp[i1][j][num(i, i1)], dp[i][j1][num(j1, j)]) |
| 14 | + */ |
| 15 | + const cache = {} |
| 16 | + const solve = (i, j, k) => { |
| 17 | + console.assert(i >= 0 && i < n) |
| 18 | + console.assert(j >= 0 && j < n) |
| 19 | + if (cache[`${i}_${j}_${k}`] != null) { |
| 20 | + return cache[`${i}_${j}_${k}`] |
| 21 | + } |
| 22 | + while (v[i] === k) i++ |
| 23 | + while (v[j] === k) j-- |
| 24 | + if (i > j) return 0 |
| 25 | + const k1 = v[i]; const k2 = v[j] |
| 26 | + const d1 = Math.abs(k1 - k); const d2 = Math.abs(k2 - k) |
| 27 | + const ans = Math.min(solve(i, j, k1) + Math.min(d1, d - d1), solve(i, j, k2) + Math.min(d2, d - d2)) |
| 28 | + if (ans < 0) console.log(i, j, k, k1, k2, d, d1, d2) |
| 29 | + return cache[`${i}_${j}_${k}`] = ans |
| 30 | + } |
| 31 | + console.log(`Case #${kase}: ${solve(0, n - 1, 0)}`) |
| 32 | + } |
| 33 | +}) |
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