并查集（洛谷试炼场——提高历练地）
并查集
普通并查集
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查找+路径压缩
int find(int x){ if(pre[x] == x) return x; return pre[x] = find(pre[x]); } -
合并
void Union(int x, int y){ int fx = find(x), fy = find(y); if(fx != fy) pre[fx] = fy; }
例题：P3367【模板】并查集
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思路：
#include#define rep(i, a, b) for(int i = (a); i <= (b); i++) #define per(i, a, b) for(int i = (a); i >= (b); i--) #define ll long long #define db double #define PII pair<int, int> #define pi acos(-1.0) const int INF = 0x7fffffff; const int N = 1e5 + 5; const db eps = 1e-10; using namespace std; int n, m, pre[N]; int find(int x){ if(pre[x] == x) return x; return pre[x] = find(pre[x]); } void join(int x,int y){ int fx = find(x), fy = find(y); if(fx != fy) pre[fx] = fy; } int main(){ cin >> n >> m; rep(i, 1, n) pre[i] = i; rep(i, 1, m){ int op, x, y; scanf("%d%d%d", &op, &x, &y); if(op == 1) join(x, y); else puts(find(x) == find(y) ? "Y" : "N"); } }
例题：P1111 修复公路
- 思路：
#include#define rep(i, a, b) for(int i = (a); i <= (b); i++) #define per(i, a, b) for(int i = (a); i >= (b); i--) #define ll long long #define db double #define PII pair<int, int> #define pi acos(-1.0) const int INF = 0x7fffffff; const int N = 1e3 + 5, M = 1e5 + 5; const db eps = 1e-10; using namespace std; int m, n, pre[N], now = 0, cnt = 0; struct AC{ int x, y, t; }a[M]; bool cmp(AC a, AC b){ return a.t < b.t; } int find(int x){ if(pre[x] == x) return x; return pre[x] = find(pre[x]); } void join(int x, int y){ int fx = find(x), fy = find(y); if(fy != fx) pre[fx] = fy; } int main(){ cin >> n >> m; rep(i, 1, m) scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].t); sort(a + 1, a + 1 + m, cmp); rep(i, 1, n) pre[i] = i; rep(i, 1, m){ int nx = a[i].x, ny = a[i].y; if(find(nx) != find(ny)){ //只统计可连接不同集合内的 join(nx, ny); now = a[i].t; cnt++; } if(cnt == n - 1){ //n个节点只需要 n - 1 条连接两个不同集合内的边即可全部连接 cout << now << endl; return 0; } } if(cnt < n - 1) cout << -1 << endl; }
例题：P1197星球大战
- 思路：
//并查集 && 逆推思想 #include#define rep(i, a, b) for(int i = (a); i <= (b); i++) #define per(i, a, b) for(int i = (a); i >= (b); i--) #define ll long long #define db double #define PII pair<int, int> #define pi acos(-1.0) const int INF = 0x7fffffff; const int N = 4e5 + 5; const db eps = 1e-10; using namespace std; int m, n, k, a[N], pre[N], ans[N]; bool vis[N]; vector<int> g[N]; int find(int x){ if(pre[x] == x) return x; return pre[x] = find(pre[x]); } int main(){ cin >> n >> m; rep(i, 1, m){ int x, y; scanf("%d%d", &x, &y); g[x].push_back(y), g[y].push_back(x); } cin >> k; rep(i, 0, n - 1) vis[i] = 1, pre[i] = i; rep(i, 1, k){ scanf("%d", &a[i]); vis[a[i]] = 0; } ans[k + 1] = n - k; rep(i, 0, n - 1){ //初始化 if(!vis[i]) continue; for(auto nxt : g[i]){ if(!vis[nxt]) continue; int fx = find(nxt), fy = find(i); //Union if(fx != fy){ pre[fx] = fy; ans[k + 1]--; } } } per(i, k, 1){ //逆推 int u = a[i]; vis[u] = 1, ans[i] = ans[i + 1] + 1; //加入一个a[i]的连通块 for(auto nxt : g[u]){ if(!vis[nxt]) continue; int fx = find(nxt), fy = find(u); if(fx != fy){ pre[fx] = fy; ans[i]--; } } } rep(i, 1, k + 1) cout << ans[i] << endl; }
带权并查集
v a l [ i ] val[i] val[i]记录节点 i i i与父节点的关系
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查找+路径压缩
int find(int x){ if(x == pre[x]) return x; int root = find(pre[x]); //一直找到根节点 ...//压缩时要更新权值,此时val[pre[x]]在上一行已更新完 return pre[x] = root; } -
合并
void Union(int x, int y){ int fx = find(x), fy = find(y); if(fx != fy){ pre[fx] = fy; ...//更新权值 } }
例题：P1196银河英雄传说
- 思路：
#include#define rep(i, a, b) for(int i = (a); i <= (b); i++) #define per(i, a, b) for(int i = (a); i >= (b); i--) #define ll long long #define db double #define PII pair<int, int> #define pi acos(-1.0) const int INF = 0x7fffffff; const int N = 3e4 + 5; const db eps = 1e-10; using namespace std; int t, n, a[N], pre[N], val[N], sz[N], x, y; //sz数组记录所在列的大小, val[i]数组记录 i 到本列根节点的距离 char op; int find(int x){ if(x == pre[x]) return x; int fx = find(pre[x]); val[x] = val[pre[x]] + val[x]; //val[x] 表示 x->pre[x], val[fx] 表示 pre[x]->fx. 上一行函数已更新 val[pre[x]] sz[x] = sz[fx]; return pre[x] = fx; } void Union(int x, int y){ int fx = find(x), fy = find(y); pre[fx] = fy; val[fx] += sz[fy]; sz[fy] = sz[fx] = sz[fx] + sz[fy]; } int main(){ cin >> t; rep(i, 1, N) pre[i] = i, val[i] = 0, sz[i] = 1; rep(i, 1, t){ cin >> op >> x >> y; if(op == 'M') Union(x, y); else{ if(find(x) == find(y)) cout << abs(val[x] - val[y]) - 1 << endl; else cout << -1 << endl; } rep(i, 1, 4) cout << i << " " << pre[i] << " " << val[i] << endl; } }
种类并查集
种类并查集一般可以用扩展域并查集，也就是增加维度，每个元素的每个维度都是一个节点，判断两个节点是否在同一个集合。
例题：P2024食物链
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思路：
//扩展域并查集 #include#define rep(i, a, b) for(int i = (a); i <= (b); i++) #define per(i, a, b) for(int i = (a); i >= (b); i--) #define ll long long const int INF = 0x7fffffff; const int N = 3e5 + 5; using namespace std; int k, n, pre[N], op[N], x[N], y[N], res = 0; int find(int x){ if(pre[x] == x) return x; return pre[x] = find(pre[x]); } void Union(int x, int y){ int fx = find(x), fy = find(y); if(fx != fy) pre[fx] = fy; } int main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); cin >> n >> k; rep(i, 1, 3 * n) pre[i] = i; rep(i, 1, k){ cin >> op[i] >> x[i] >> y[i]; if(x[i] > n || y[i] > n || (op[i] == 2 && x[i] == y[i])){ res++, cout << i << endl; continue; } int xself = x[i], xenemy = x[i] + n, xeat = x[i] + n + n; int yself = y[i], yenemy = y[i] + n, yeat = y[i] + n + n; if(op[i] == 1){ if(find(xself) == find(yeat) || find(xself) == find(yenemy)) res++, cout << i << endl; else Union(xself, yself), Union(xenemy, yenemy), Union(xeat, yeat); } else{ if(find(xself) == find(yself) || find(xself) == find(yeat)) res++, cout << i << endl; else Union(xself, yenemy), Union(xenemy, yeat), Union(xeat, yself); } } cout << res << endl; } 当然用带权并查集也能做
//带权并查集 #include#define rep(i, a, b) for(int i = (a); i <= (b); i++) #define per(i, a, b) for(int i = (a); i >= (b); i--) #define ll long long const int INF = 0x7fffffff; const int N = 1e5 + 5; using namespace std; int k, n, pre[N], op[N], x[N], y[N], res = 0, val[N]; int find(int x){ if(pre[x] == x) return x; int root = find(pre[x]); (val[x] += val[pre[x]]) %= 3; //更新权值 return pre[x] = root; } int main(){ ios::sync_with_stdio(0), cin.tie(0); cin >> n >> k; rep(i, 1, n) pre[i] = i, val[i] = 0; rep(i, 1, k){ cin >> op[i] >> x[i] >> y[i]; if(x[i] > n || y[i] > n || (op[i] == 2 && x[i] == y[i])){ res++; continue; } int fx = find(x[i]), fy = find(y[i]); if(fx == fy){ if(op[i] == 1 && val[x[i]] != val[y[i]]) res++; if(op[i] == 2 && val[x[i]] != (val[y[i]] + 1) % 3) res++; } else{ pre[fx] = fy; val[fx] = (val[y[i]] + (op[i] == 1 ? 0 : 1) - val[x[i]] + 3) % 3; } } cout << res << endl; } //带权并查集 #include#define rep(i, a, b) for(int i = (a); i <= (b); i++) #define per(i, a, b) for(int i = (a); i >= (b); i--) #define ll long long const int INF = 0x7fffffff; const int N = 1e5 + 5; using namespace std; int k, n, pre[N], op[N], x[N], y[N], res = 0, val[N]; int find(int x){ if(pre[x] == x) return x; int root = find(pre[x]); (val[x] += val[pre[x]]) %= 3; //更新权值 return pre[x] = root; } void Union(int x, int y, int op){ int fx = find(x), fy = find(y); if(fx != fy){ pre[fx] = fy; val[fx] = (val[y] + (op == 1 ? 0 : 1) - val[x] + 3) % 3; } } int main(){ ios::sync_with_stdio(0), cin.tie(0); cin >> n >> k; rep(i, 1, n) pre[i] = i, val[i] = 0; rep(i, 1, k){ cin >> op[i] >> x[i] >> y[i]; if(x[i] > n || y[i] > n || (op[i] == 2 && x[i] == y[i])){ res++; continue; } if(op[i] == 1){ if(find(x[i]) == find(y[i]) && val[x[i]] != val[y[i]]) res++; else Union(x[i], y[i], op[i]); } else{ if(find(x[i]) == find(y[i]) && val[x[i]] != (val[y[i]] + 1) % 3) res++; else Union(x[i], y[i], op[i]); } } cout << res << endl; }