HDOJ 1512 几乎模板的左偏树
题目大意：有n个猴子。每个猴子有一个力量值，力量值越大表示这个猴子打架越厉害。如果2个猴子不认识，他们就会找他们认识的猴子中力量最大的出来单挑，单挑不论输赢，单挑的2个猴子力量值减半，这2拨猴子就都认识了，不打不相识嘛。现在给m组询问，如果2只猴子相互认识，输出-1，否则他们各自找自己认识的最牛叉的猴子单挑，求挑完后这拨猴子力量最大值。
左偏大根加并查
代码都懒的贴了...
1 #include <iostream> 2 #include <fstream> 3 #include <sstream> 4 #include <algorithm> 5 #include <string> 6 #include <set> 7 #include <map> 8 #include <utility> 9 #include <queue> 10 #include <stack> 11 #include <list> 12 #include <vector> 13 #include <cstdio> 14 #include <cstdlib> 15 #include <cstring> 16 #include <cmath> 17 #include <ctime> 18 using namespace std; 19 const int maxn = 100010; 20 struct node { 21 int l,r,dis,val,par; 22 } heap[maxn]; 23 int N, M; 24 int find ( int &x ) { 25 return heap[x].par == x ? x : heap[x].par = find ( heap[x].par ); 26 } 27 int merge(int rt1,int rt2) 28 { 29 if (rt1==0) return rt2; 30 if (rt2==0) return rt1; 31 if ( heap[rt2].val>heap[rt1].val )swap(rt1,rt2); 32 heap[rt1].r = merge(heap[rt1].r,rt2); 33 heap[heap[rt1].r].par = rt1; 34 if ( heap[ heap[rt1].l ].dis < heap[ heap[rt2].r ].dis ) 35 swap ( heap[rt1].l, heap[rt1].r ); 36 else heap[rt1].dis = heap[ heap[rt1].r ].dis + 1; 37 return rt1; 38 } 39 int push ( int x, int y ) { 40 return merge ( x, y ); 41 } 42 int pop ( int &x ) { 43 int l = heap[x].l; 44 int r = heap[x].r; 45 heap[l].par = l; 46 heap[r].par = r; 47 heap[x].l = heap[x].r = heap[x].dis = 0; 48 return merge ( l, r ); 49 } 50 bool scan_d(int &num) { 51 char in;bool IsN=false; 52 in=getchar(); 53 if(in==EOF) return false; 54 while(in!='-'&&(in<'0'||in>'9')) in=getchar(); 55 if(in=='-'){ IsN=true;num=0;} 56 else num=in-'0'; 57 while(in=getchar(),in>='0'&&in<='9'){ 58 num*=10,num+=in-'0'; 59 } 60 if(IsN) num=-num; 61 return true; 62 } 63 int main() { 64 while ( scan_d ( N ) ) { 65 for ( int i = 1; i <= N; ++ i ) { 66 scan_d ( heap[i].val ); 67 heap[i].l = heap[i].r = heap[i].dis = 0; 68 heap[i].par = i; 69 } 70 scan_d ( M ); 71 int a, b, x, y; 72 while ( M -- ) { 73 scan_d (a); scan_d (b); 74 x = find ( a ); 75 y = find ( b ); 76 if ( x == y ) { 77 puts ( "-1" ); 78 } else { 79 heap[x].val /= 2; 80 int px = push ( pop ( x ), x ); 81 heap[y].val /= 2; 82 int py = push ( pop ( y ), y ); 83 84 printf ( "%d\n", heap[ merge ( px, py ) ].val ); 85 } 86 } 87 } 88 return 0; 89 }
