zoj 2334 Monkey King/左偏树+并查集

来源：互联网发布：战斗力获取数据超时编辑：程序博客网时间：2024/05/16 13:03

原题链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1389

大致题意：N只相互不认识的猴子(每只猴子有一个战斗力值)

两只不认识的猴子之间发生冲突，两只猴子会分别请出它们认识的最强壮的猴子进行决斗。

决斗之后这，两群猴子都相互认识了。决斗的那两只猴子战斗力减半。。。

有m组询问，输入a b表示猴子a和b发生了冲突，若a,b属于同一个集合输出-1

否则输出决斗之后这群猴子(已合并)中最强的战斗力值。。。

具体思路：用并查集判断是否属于同一集合，用左偏树维护一群猴子的战斗力值。

加了垃圾回收，否则容易爆内存。。。

具体如下：

#include<cstdio>#include<cstdlib>#include<iostream>#include<algorithm>const int Max_N = 100050;struct UnionFind{int par[Max_N], rank[Max_N];inline void init(int n){for (int i = 1; i <= n; i++){par[i] = i;rank[i] = 0;}}inline int find(int x){while (x != par[x]){x = par[x] = par[par[x]];}return x;}inline void unite(int x, int y){x = find(x), y = find(y);if (x == y) return;if (rank[x] < rank[y]){par[x] = y;} else {par[y] = x;if (rank[x] == rank[y]) rank[x]++;}}};struct Node{int v, npl;Node *ch[2];inline void set(int _v = 0, int _npl = -1, Node *p = NULL){v = _v, npl = _npl;ch[0] = ch[1] = p;}inline void push_up(){npl = ch[1]->npl + 1;}};struct LeftistTree{int N, top;UnionFind rec;Node *tail, *null;Node stack[Max_N], *ptr[Max_N], *store[Max_N];void init(int n){tail = &stack[0];null = tail++;null->set();N = n, top = 0, rec.init(n);}inline Node *newNode(int v){Node *p = null;if (!top) p = tail++;else p = store[--top];p->set(v, 0, null);return p;}inline Node* Merge(Node* &x, Node* &y){if (x == null) return y;if (y == null) return x;if (y->v > x->v) std::swap(x, y);x->ch[1] = Merge(x->ch[1], y);if (x->ch[1]->npl > x->ch[0]->npl) std::swap(x->ch[0], x->ch[1]);x->push_up();return x;}inline int get_max(int i){return ptr[i]->v;}inline void insert(){int v;for (int i = 1; i <= N; i++){scanf("%d", &v);ptr[i] = newNode(v);}}inline void del(int i){int ret = get_max(i);Node *x = newNode(ret >> 1);store[top++] = ptr[i];ptr[i] = Merge(ptr[i]->ch[0], ptr[i]->ch[1]);ptr[i] = Merge(ptr[i], x);}inline void gogo(int a, int b){int ans = 0;a = rec.find(a), b = rec.find(b);if (a == b){printf("-1\n");return;}rec.unite(a, b);del(a), del(b);if (rec.rank[a] > rec.rank[b]){ptr[a] = Merge(ptr[a], ptr[b]);ans = ptr[a]->v;} else {ptr[b] = Merge(ptr[a], ptr[b]);ans = ptr[b]->v;}printf("%d\n", ans);}}lft;int main(){#ifdef LOCALfreopen("in.txt", "r", stdin);freopen("out.txt", "w+", stdout);#endifint n, m, a, b;while (~scanf("%d", &n)){lft.init(n), lft.insert();scanf("%d", &m);while (m--){scanf("%d %d", &a, &b);lft.gogo(a, b);}}return 0;}

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