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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:
1 #include<cstdio> 2 #include<cstdlib> 3 #include<iostream> 4 #include<algorithm> 5 const int Max_N = 100050; 6 struct UnionFind{ 7 int par[Max_N], rank[Max_N]; 8 inline void init(int n){ 9 for (int i = 1; i <= n; i++){ 10 par[i] = i; 11 rank[i] = 0; 12 } 13 } 14 inline int find(int x){ 15 while (x != par[x]){ 16 x = par[x] = par[par[x]]; 17 } 18 return x; 19 } 20 inline void unite(int x, int y){ 21 x = find(x), y = find(y); 22 if (x == y) return; 23 if (rank[x] < rank[y]){ 24 par[x] = y; 25 } else { 26 par[y] = x; 27 if (rank[x] == rank[y]) rank[x]++; 28 } 29 } 30 }; 31 struct Node{ 32 int v, npl; 33 Node *ch[2]; 34 inline void set(int _v = 0, int _npl = -1, Node *p = NULL){ 35 v = _v, npl = _npl; 36 ch[0] = ch[1] = p; 37 } 38 inline void push_up(){ 39 npl = ch[1]->npl + 1; 40 } 41 }; 42 struct LeftistTree{ 43 int N, top; 44 UnionFind rec; 45 Node *tail, *null; 46 Node stack[Max_N], *ptr[Max_N], *store[Max_N]; 47 void init(int n){ 48 tail = &stack[0]; 49 null = tail++; 50 null->set(); 51 N = n, top = 0, rec.init(n); 52 } 53 inline Node *newNode(int v){ 54 Node *p = null; 55 if (!top) p = tail++; 56 else p = store[--top]; 57 p->set(v, 0, null); 58 return p; 59 } 60 inline Node* Merge(Node* &x, Node* &y){ 61 if (x == null) return y; 62 if (y == null) return x; 63 if (y->v > x->v) std::swap(x, y); 64 x->ch[1] = Merge(x->ch[1], y); 65 if (x->ch[1]->npl > x->ch[0]->npl) 66 std::swap(x->ch[0], x->ch[1]); 67 x->push_up(); 68 return x; 69 } 70 inline int get_max(int i){ 71 return ptr[i]->v; 72 } 73 inline void insert(){ 74 int v; 75 for (int i = 1; i <= N; i++){ 76 scanf("%d", &v); 77 ptr[i] = newNode(v); 78 } 79 } 80 inline void del(int i){ 81 int ret = get_max(i); 82 Node *x = newNode(ret >> 1); 83 store[top++] = ptr[i]; 84 ptr[i] = Merge(ptr[i]->ch[0], ptr[i]->ch[1]); 85 ptr[i] = Merge(ptr[i], x); 86 } 87 inline void gogo(int a, int b){ 88 int ans = 0; 89 a = rec.find(a), b = rec.find(b); 90 if (a == b){ 91 printf("-1\n"); 92 return; 93 } 94 rec.unite(a, b); 95 del(a), del(b); 96 if (rec.rank[a] > rec.rank[b]){ 97 ptr[a] = Merge(ptr[a], ptr[b]); 98 ans = ptr[a]->v; 99 } else { 100 ptr[b] = Merge(ptr[a], ptr[b]); 101 ans = ptr[b]->v; 102 } 103 printf("%d\n", ans); 104 } 105 }lft; 106 int main(){ 107 #ifdef LOCAL 108 freopen("in.txt", "r", stdin); 109 freopen("out.txt", "w+", stdout); 110 #endif 111 int n, m, a, b; 112 while (~scanf("%d", &n)){ 113 lft.init(n), lft.insert(); 114 scanf("%d", &m); 115 while (m--){ 116 scanf("%d %d", &a, &b); 117 lft.gogo(a, b); 118 } 119 } 120 return 0; 121 }
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原文地址:http://www.cnblogs.com/GadyPu/p/4463978.html