[AcWing Base course] DSA/UnionFind

Author: houhuawei23

DSA/UnionFind/AW240.cpp

@@ -0,0 +1,114 @@
+/*
+AcWing
+240. 食物链
+https://www.acwing.com/problem/content/242/
+
+if x and y are in same set(px == py), we can determine their relation by d.
+d[i] record the distance from i to the root of the set tree
+    (distance of the food chain, not the set tree height)
+- if (d[x] - d[y]) % 3 == 0, same kind
+- if ((d[x] - d[y]) % 3 + 3) % 3 == 1, x eat y
+
+if x and y not in the same set (px != py), we need to merge two set tree:
+- p[px] = py: set father of px to be py
+- determine the 'd' of px
+    - if x, y are same kind:
+        d[px] = d[y] - d[x]
+        because x and y are same kind, so
+        - d[x] -> d[y],
+        - d[px] -> new d[x] - old d[x] = d[y] - d[x]
+    - if x eat y:
+        we need to insert x under y, so
+        - d[x] -> d[y] + 1
+        - d[px] -> new d[x] - old d[x] = d[y] + 1 - d[x]
+    we only need to update d of px, because when we update d of px,
+    the followings under px will be auto update by the call of find()
+
+because union-find set need to flatten the set tree,
+so we need to update the d of each node when calling find()
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 5e4 + 10;
+const int K = 1e5 + 10;
+
+int n, k;
+int cnt;
+
+int p[N], d[N];
+
+int find(int x) {
+    if(p[x] != x) {
+        int t = find(p[x]);
+        d[x] += d[p[x]]; /* old p[x] */
+        p[x] = t;
+    }
+    return p[x];
+}
+
+int main() {
+    cin >> n >> k;
+    for(int i = 1; i <= n; i++) {
+        p[i] = i;
+    }
+    while(k--) {
+        int D, x, y;
+        cin >> D >> x >> y;
+        if(x > n or y > n) {
+            cnt++;
+            continue;
+        }
+        int px = find(x), py = find(y);
+
+        if(D == 1) {
+            /* say x, y 同类 */
+            if(px == py) { /* smae set, can determin relation */
+                if((d[x] - d[y]) % 3 != 0) {
+                    cnt++;
+                    continue;
+                }
+            } else {
+                p[px] = py;          /* merge px to py */
+                d[px] = d[y] - d[x]; /* same class */
+            }
+
+        } else if(D == 2) {
+            /* say x eat y */
+            if(x == y) {
+                cnt++;
+                continue;
+            }
+            if(px == py) {
+                /* in same set, known the relation */
+                // if((d[x] - d[y] - 1) % 3)
+                if(((d[x] - d[y]) % 3 + 3) % 3 != 1) {
+                    cnt++;
+                    continue;
+                }
+            } else {
+                /* not int samse set, merge and build relation (d) */
+                /* actrualy, we insert x under y,
+                ** so the d of px = d[y] + 1 - d[x] */
+                p[px] = py; /* merge px to py */
+                d[px] = d[y] + 1 - d[x];
+            }
+        } else {
+            assert(false);
+        }
+    }
+    cout << cnt << endl;
+}
+/*
+输入样例：
+100 7
+1 101 1
+2 1 2
+2 2 3
+2 3 3
+1 1 3
+2 3 1
+1 5 5
+输出样例：
+3
+*/

DSA/UnionFind/AW240_noted.cpp

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+/*
+作者：yxc
+链接：https://www.acwing.com/activity/content/code/content/45325/
+来源：AcWing
+著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
+*/
+
+#include <iostream>
+
+using namespace std;
+
+const int N = 1e5 + 10;
+
+int n, m;
+int p[N], d[N];  // p[]寻找祖宗节点，d[]求到祖宗节点的距离
+
+int find(int x) {
+    if(p[x] != x) {
+        int t = find(p[x]);  // u暂时存一下p[x]根节点，辅助变量
+        d[x] += d[p[x]];     // 更新距离
+        p[x] = t;
+    }
+    return p[x];
+}
+
+// 不行，因为这个路径的长度（高度），是需要自上而下加起来的，从根节点往下走
+// 所以要先调用递归
+
+// int find(int x)
+// {
+//     if (p[x] != x)
+//     {
+//         d[x] += d[p[x]];    // 更新距离
+//         p[x] = find(p[x]);
+//     }
+//     return p[x];
+// }
+
+int main() {
+    scanf("%d%d", &n, &m);
+
+    for(int i = 1; i <= n; i++)
+        p[i] = i;
+
+    int res = 0;  // 记录错误数
+    while(m--) {
+        int t, x, y;
+        scanf("%d%d%d", &t, &x, &y);
+
+        if(x > n || y > n)
+            res++;  // 当前的话中X或Y比N大，是假话
+        else {
+            int px = find(x), py = find(y);  // 查找根节点
+            if(t == 1) {
+                // 判断是否同类
+                if(px == py) {  // 若 x 与 y 在同一个集合中
+                    if((d[x] - d[y]) % 3)
+                        res++;
+                    // 两数到根节点距离之差的模不为
+                    // 0，说明不是同一类，是假话
+                    // 其中 (d[x] - d[y]) % 3 不可写为 d[x] % 3 != d[y] % 3
+                    // 因为 d[x], d[y] 可能为负数（一正一负），可改做 (d[x] % 3
+                    // + 3) % 3 != (d[y] % 3 + 3) % 3 负数 mod 正数为负数
+                } else {         // 则 x 与 y 不在同一个集合中
+                    p[px] = py;  // x 所在集合 合并到 y 所在集合
+                    d[px] = d[y] - d[x];
+                    // d[x] 的距离为什么不更新？
+                    // 只是暂时不更新，在调用 find 时再更新
+                }
+            } else {
+                // X 是否吃 Y
+                if(px == py) {
+                    // 若 x 与 y 在同一个集合中
+                    // 若 X 吃 Y，则 d[x] 比 d[y] 大 1
+                    if((d[x] - d[y] - 1) % 3)
+                        res++;
+                    // 若距离之差 - 1 的模不为 0，说明吃不掉，是假话
+                } else {
+                    // 则 x 与 y 不在同一个集合中
+                    p[px] = py; /* 合并集合 */
+                    // (d[x] - d[y] - 1) % 3 == 0
+                    // d[x] + d[px] - 1 = d[y]  则：
+                    d[px] = d[y] + 1 - d[x];
+                }
+            }
+        }
+    }
+
+    printf("%d\n", res);
+
+    return 0;
+}

DSA/UnionFind/AW837.cpp

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+/*
+AcWing 837. 连通块中点的数量
+https://www.acwing.com/problem/content/839/
+输入样例：
+5 5
+C 1 2
+Q1 1 2
+Q2 1
+C 2 5
+Q2 5
+输出样例：
+Yes
+2
+3
+*/
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1e5 + 10;
+int n, m;
+int p[N], cnt[N];
+
+int find(int x) {
+    if(p[x] != x) {
+        p[x] = find(p[x]);
+    }
+    return p[x];
+}
+
+int main() {
+    cin >> n >> m;
+    for(int i = 1; i <= n; i++) {
+        p[i] = i;
+        cnt[i] = 1;
+    }
+    string op;
+    int a, b;
+    while(m--) {
+        cin >> op;
+        if(op == "C") {
+            cin >> a >> b;
+            int pa = find(a), pb = find(b);
+            if(pa != pb) {
+                p[pa] = pb;
+                cnt[pb] += cnt[pa];
+            }
+        } else if(op == "Q1") {
+            cin >> a >> b;
+            int pa = find(a), pb = find(b);
+            if(pa == pb) {
+                cout << "Yes" << endl;
+            } else {
+                cout << "No" << endl;
+            }
+        } else if(op == "Q2") {
+            cin >> a;
+            int pa = find(a);
+            cout << cnt[pa] << endl;
+        }
+    }
+}

Math/testmod.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    cout << (-2) % 3 << endl;
+    cout << (-3) % 3 << endl;
+    cout << (6) % 3 << endl; /* 0 */
+    cout << (-1) % 3 << endl; 
+    cout << (-2) % 3 << endl;
+}