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刷题/Codeforces/DP/Codeforces - 977F. Consecutive Subsequence( Map + DP) & 1097B. Petr and a Combination Lock(枚举).md → Algorithm/Codeforces/DP/Codeforces - 977F. Consecutive Subsequence( Map + DP) & 1097B. Petr and a Combination Lock(枚举).md

@@ -4,13 +4,13 @@
 * Codeforces - 1097B. Petr and a Combination Lock(枚举)
 ***
 
-### <font color = red id = "1">Codeforces - 977F. Consecutive Subsequence( Map + DP)
+### Codeforces - 977F. Consecutive Subsequence( Map + DP)
 #### [题目链接](https://codeforces.com/problemset/problem/977/F)
 
 > https://codeforces.com/problemset/problem/977/F
 
 #### 题目
-**求最长的连续递增子序列(注意是<font color = blue>连续递增</font>(也就是值前后相差为`1`))， 输出长度以及子序列的下标。**
+**求最长的连续递增子序列(注意是连续递增(也就是值前后相差为`1`))， 输出长度以及子序列的下标。**
 
 ![在这里插入图片描述](images/977F_t.png)
 ![在这里插入图片描述](images/977F_t2.png)

Algorithm/Codeforces/DataStructure/Map/Codeforces - 1133D. Zero Quantity Maximization.md

@@ -0,0 +1,106 @@
+## Codeforces - 1133D. Zero Quantity Maximization(哈希)
+
+#### [题目链接](https://codeforces.com/problemset/problem/1133/D)
+
+> https://codeforces.com/problemset/problem/1133/D
+
+#### 题目
+
+就是给你`n`，然后两个数目为`n`个数组`a[]、b[]`，要你随便找一个数`d`（可以是浮点数和整数），使得`d * a[i] + b[i] = 0`的数目数目最多。求出最多的`0`。
+
+![1133D_t.png](images/1133D_t.png)
+
+![1133D_t2.png](images/1133D_t2.png)
+
+### 解析
+
+重点是怎么哈希，这里先求得两个数的最大公约数，然后相除，然后在`map`中的数只要是`a == a2 && b == b2`或者`a == -a2 && b == -b2`的就可以看做同一个。
+
+一开始没有注意`if(a[i] == 0 && b[i] != 0) continue;`，在这里wrong了挺多次。这种情况是不可能为`0`的。
+
+```java
+import java.io.*;
+import java.util.*;
+
+public class Main {
+
+    static PrintStream out = System.out;
+
+    static class Pair {
+        int f, s;
+
+        public Pair(int f, int s) {
+            this.f = f;
+            this.s = s;
+        }
+
+        @Override
+        public boolean equals(Object o) {
+            Pair tr = (Pair) o;
+            return (f == tr.f && s == tr.s) || (f == -tr.f && s == -tr.s);
+        }
+        @Override
+        public int hashCode() {
+            return 31 * f + s;
+        }
+    }
+
+    // 12 16
+    static int gcd(int a, int b) {
+        int r = 0;
+        while (b != 0) {
+            r = a % b;
+            a = b;
+            b = r;
+        }
+        return a;
+    }
+
+    static void solve(InputStream stream) {
+        Scanner in = new Scanner(new BufferedInputStream(stream));
+        int n = in.nextInt();
+        int[] a = new int[n];
+        int[] b = new int[n];
+        boolean good = false;
+        for (int i = 0; i < n; i++) {
+            a[i] = in.nextInt();
+            if (a[i] != 0) good = true;
+        }
+        for (int i = 0; i < n; i++) {
+            b[i] = in.nextInt();
+            if (b[i] == 0) good = true;
+        }
+        if (!good) {
+            System.out.println(0);
+            return;
+        }
+        HashMap<Pair, Integer> map = new HashMap<>();
+        int allZero = 0, fiZero = 0;
+        for (int i = 0; i < n; i++) {
+            if(a[i] == 0 && b[i] != 0) continue;//在这里wrong了挺多次
+            if(a[i] != 0 && b[i] == 0)  {
+                fiZero++;
+                continue;
+            }
+            int d = gcd(a[i], b[i]);
+            if (d != 0) {
+                Pair p = new Pair(a[i] / d, b[i] / d);
+                map.put(p, map.getOrDefault(p, 0) + 1);
+            } else {
+                allZero++;
+            }
+        }
+        int res = fiZero; // 有可能是 0 -> X ，就直接取0了
+        for (Pair key : map.keySet())
+            res = Math.max(res, map.get(key));
+        out.println(res + allZero);
+    }
+
+    public static void main(String[] args) {
+        solve(System.in);
+    }
+}
+```
+
+
+

刷题/Codeforces/Graph/Codeforces - 1082A & 1073A & 330B & GYM101502I(最短路).md → Algorithm/Codeforces/Graph/Codeforces - 1082A & 1073A & 330B & GYM101502I(最短路).md

@@ -6,7 +6,7 @@
 * [Codeforces - 330B - Road Construction](#codeforces---330b---road-construction)
 * [GYM - 101502I - Move Between Numbers](#gym---101502i---move-between-numbers)
 ***
-### <font color = red id = "1">Codeforces - 1082A - Vasya and Book
+### Codeforces - 1082A - Vasya and Book
 #### [题目链接](http://codeforces.com/problemset/problem/1082/A)
 
 > http://codeforces.com/problemset/problem/1082/A

Algorithm/Codeforces/Graph/Codeforces - 1133F1. Spanning Tree with Maximum Degree.md

@@ -0,0 +1,90 @@
+## Codeforces - 1133F1. Spanning Tree with Maximum Degree.md(DFS，记录父亲)
+
+#### [题目链接](https://codeforces.com/problemset/problem/1133/F1)
+
+> https://codeforces.com/problemset/problem/1133/F1
+
+#### 题目
+
+给你一个图，要你求必须包含**度最大的点的所有相连边**的生成树。
+
+![1133F1_t.png](images/1133F1_t.png)
+
+![1133F1_t2.png](images/1133F1_t2.png)
+
+![1133F1_t3.png](images/1133F1_t3.png)
+
+![1133F1_t4.png](images/1133F1_t4.png)
+
+![1133F1_t5.png](images/1133F1_t5.png)
+
+#### 解析
+
+思路：
+
+* 先找出度最大的度的顶点`maxv`。然后先将这个点的邻接边加入结果。
+* 然后从和`maxv`邻接的点开始`dfs`，这个过程既要记录`vis`是否访问，又要记录一个父亲节点，这样才比较好维护；
+
+![1133F1_s.png](images/1133F1_s.png)
+
+代码:
+
+```java
+import java.io.*;
+import java.util.*;
+
+public class Main {
+
+    static PrintStream out = System.out;
+
+    static boolean[] vis;
+    static ArrayList<Integer> G[];
+
+    static void solve(InputStream stream) {
+        Scanner in = new Scanner(new BufferedInputStream(stream));
+        int n = in.nextInt();
+        int m = in.nextInt();
+        G = new ArrayList[n + 1];
+        int[] deg = new int[n + 1];
+        for (int i = 1; i <= n; i++) G[i] = new ArrayList<>();
+        for (int i = 0; i < m; i++) {
+            int from = in.nextInt();
+            int to = in.nextInt();
+            deg[to]++;
+            deg[from]++;
+            G[from].add(to);
+            G[to].add(from);
+        }
+        int maxDeg = 0;
+        int maxv = 0;
+        for (int i = 1; i <= n; i++) { // 找度最大的
+            if (deg[i] > maxDeg) {
+                maxDeg = deg[i];
+                maxv = i;
+            }
+        }
+        vis = new boolean[n + 1];
+        vis[maxv] = true;
+        for(int to : G[maxv]){ // 先将度最大的相连的边连接起来
+            vis[to] = true;
+            out.println(maxv + " " + to);
+        }
+        for(int to : G[maxv]) dfs(to, maxv);
+    }
+
+    static void dfs(int cur, int par) { // 记录当前节点和父亲节点
+//        if(vis[cur]) return ; // 不要写这里，不然第一次就不能进去
+        vis[cur] = true;
+        for (int to : G[cur]) {
+            if(to == par || vis[to]) continue;
+            out.println(cur + " " + to);
+            dfs(to, cur);
+        }
+    }
+
+    public static void main(String[] args) {
+        solve(System.in);
+    }
+}
+```
+

Algorithm/Codeforces/Greedy/Codeforces - 1130B. Two Cakes.md

@@ -0,0 +1,72 @@
+## Codeforces - 1130B. Two Cakes
+
+#### [题目链接](https://codeforces.com/problemset/problem/1130/B)
+
+> https://codeforces.com/problemset/problem/1130/B
+
+#### 题目
+
+给你`n`和`2n`个数。(`1~n`之间的数各两个)
+
+两个人都同时从最左边的`1`位置开始出发，每次依次去找`1, 2, 3, ..., n`。问你两个人需要走的最小的路程。
+
+![1130_B_t.png](images/1130_B_t.png)
+
+![1130_B_t2.png](images/1130_B_t2.png)
+
+### 解析
+
+贪心的策略就是:
+
+当前的`abs(id1 - p1[i + 1]) + abs(id2 - p2[i + 1]) `和`abs(id1 - p2[i + 1]) + abs(id2 - p1[i + 1])`哪个更小，就去哪个。
+
+```java
+import java.io.*;
+import java.util.*;
+
+public class Main {
+
+    public static void main(String[] args) {
+        Scanner in = new Scanner(new BufferedInputStream(System.in));
+        PrintStream out = System.out;
+        int n = in.nextInt();
+        int[] arr = new int[n * 2 + 1];
+        int[] p1 = new int[n * 2 + 1];
+        int[] p2 = new int[n * 2 + 1];
+        for (int i = 1; i <= 2 * n; i++) {
+            arr[i] = in.nextInt();
+            if (p1[arr[i]] == 0)
+                p1[arr[i]] = i;
+            else
+                p2[arr[i]] = i;
+        }
+
+//        System.out.println(Arrays.toString(p1));
+//        System.out.println(Arrays.toString(p2));
+
+        int id1 = 1, id2 = 1;
+        long r1 = 0, r2 = 0;
+        for (int i = 0; i < n; i++) { // 每一个数
+            if (abs(id1 - p1[i + 1]) + abs(id2 - p2[i + 1]) <
+                    abs(id1 - p2[i + 1]) + abs(id2 - p1[i + 1])) { //贪心策略
+                r1 += abs(p1[i + 1] - id1);
+                id1 = p1[i + 1];
+                r2 += abs(p2[i + 1] - id2);
+                id2 = p2[i + 1];
+            } else {
+                r1 += abs(p2[i + 1] - id1);
+                id1 = p2[i + 1];
+                r2 += abs(p1[i + 1] - id2);
+                id2 = p1[i + 1];
+            }
+        }
+        out.println(r1 + r2);
+    }
+
+    static int abs(int n) {
+        return Math.abs(n);
+    }
+}
+
+```
+

Algorithm/Codeforces/Greedy/Codeforces - 1157B. Long Number.md

@@ -0,0 +1,48 @@
+# Codeforces - 1157B. Long Number
+
+#### [题目链接](https://codeforces.com/problemset/problem/1157/B)
+
+> https://codeforces.com/problemset/problem/1157/B
+
+#### 题目
+
+![1556624729268](assets/1556624729268.png)
+
+### 解析
+
+模拟即可。注意情况`if(f[s[i] - '0'] == s[i] - '0') continue`。
+
+```java
+import java.util.*;
+import java.io.*;
+
+public class Main{
+
+    static void solve(Scanner in, PrintWriter out){
+        int n = in.nextInt();
+        char[] s = in.next().toCharArray();
+        int[] f = new int[10];
+        for(int i = 1; i < 10; i++) f[i] = in.nextInt();
+        boolean flag = true; 
+        for(int i = 0; i < n; i++){
+            if(f[s[i] - '0'] > s[i] - '0'){
+                s[i] = (char)(f[s[i] - '0'] + '0');
+                flag = false;
+            }else {
+                if(f[s[i] - '0'] == s[i] - '0') continue; // 注意这里
+                if(!flag) break;
+            }
+        }
+        out.println(s);
+    }
+
+    public static void main(String[] args) {
+        Scanner in = new Scanner(new BufferedInputStream(System.in));
+        PrintWriter out = new PrintWriter(System.out);
+        solve(in, out);
+        out.close();
+    }
+}
+
+```
+