chore: alg

Author: ceilf6

JS/algorithm/二分答案.js

@@ -0,0 +1,93 @@
+// https://www.nowcoder.com/practice/2e27509b990d4d02a70c0f208f078cdf
+
+const rl = require("readline").createInterface({ input: process.stdin });
+var iter = rl[Symbol.asyncIterator]();
+const readline = async () => (await iter.next()).value;
+
+void async function () {
+    let [n, m] = (await readline()).split(' ').map(it => Number(it))
+    const str = await readline()
+    const Ws = []
+    let idx = str.indexOf('W', 0)
+    while (idx !== -1) {
+        Ws.push(idx);
+        idx = str.indexOf('W', idx + 1);
+    }
+
+    if (!Ws.length) {
+        console.log(0) // 注意特殊情况
+        return
+    }
+    function check(k) {
+        let idx = 0 // 当前已经覆盖第 i 个 W
+        let cnt = 0 // 已经使用了刷的次数
+        while (idx < Ws.length) {
+            cnt++
+            const curLast = Ws[idx] + k - 1
+            while (idx < Ws.length && Ws[idx] <= curLast) // 覆盖了的
+                idx++
+        }
+        return cnt <= m
+    }
+    let l = 1
+    let r = n
+    while (l <= r) {
+        const k = Math.floor((l + r) / 2)
+        if (check(k)) {
+            r = k - 1 // 尝试更小
+        } else {
+            l = k + 1
+        }
+    }
+    console.log(l)
+}()
+
+
+
+// 复杂度过高
+const rl = require("readline").createInterface({ input: process.stdin });
+var iter = rl[Symbol.asyncIterator]();
+const readline = async () => (await iter.next()).value;
+
+void async function () {
+    // 首先假如 m 只有一次，那么就是看 W 出现的两个极端位置
+    // 如果 m = 2 那么进入 W 极端位置中间找最靠近中间的
+    // m = 3 再进入
+    let [n, m] = (await readline()).split(' ').map(it => Number(it))
+    const str = await readline()
+    const Ws = []
+    let idx = str.indexOf('W', 0)
+    while (idx !== -1) {
+        Ws.push(idx);
+        idx = str.indexOf('W', idx + 1);
+    }
+    // console.log(Ws)
+    let k = Infinity;
+    // let l = 0
+    // let r = n-1
+    // while(str[l]!=='W')
+    //     l++
+    // while(str[r]!=='W')
+    //     r--
+    m -= 1
+    function dfs(m, l, r) {
+        if (m == 0) {
+            k = Math.min(k, Ws[r] - Ws[l] + 1)
+            return
+        }
+        const mid = Math.floor((l + r) / 2)
+        // let d = 0
+        // while(str[mid+d]!=='W' && str[mid+d]!=='W')
+        //     d++
+        dfs(m - 1, l, mid)
+        dfs(m - 1, mid, r)
+    }
+    dfs(m, 0, Ws.length - 1)
+    // while(m){
+    //     const mid = Math.floor((l+r)/2)
+    //     let d = 0
+    //     while(str[mid+d]!=='W')
+    //     m--
+    // }
+    console.log(k)
+}()