The main problem: there are n monkeys. Each monkey has a strength value, the greater the strength value means that the monkey fights the more severe. If 2 monkeys do not know, they will find the monkeys they know the most powerful out singled out, regardless of winning or losing, singled out the 2 monkeys power value halved, the 2 monkeys are known, do not know well. Now to the M group, if 2 monkeys know each other, output-1, or they each find their own know the most bull of the monkey singled out, to pick up after this dial the Monkey power maximum value.
Left side big root Plus and check
The code is lazy paste ...
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>Ten#include <stack> One#include <list> A#include <vector> -#include <cstdio> -#include <cstdlib> the#include <cstring> -#include <cmath> -#include <ctime> - using namespacestd; + Const intMAXN =100010; - structNode { + intL,r,dis,val,par; A } HEAP[MAXN]; at intN, M; - intFind (int&x) { - returnHeap[x].par = = x? X:heap[x].par =find (Heap[x].par); - } - intMergeintRt1,intrt2) - { in if(rt1==0)returnRt2; - if(rt2==0)returnRt1; to if(heap[rt2].val>heap[rt1].val) Swap (RT1,RT2); +HEAP[RT1].R =merge (HEAP[RT1].R,RT2); -Heap[heap[rt1].r].par =Rt1; the if(heap[heap[rt1].l].dis <heap[HEAP[RT2].R].dis) * swap (HEAP[RT1].L, HEAP[RT1].R); $ ElseHeap[rt1].dis = heap[HEAP[RT1].R].dis +1;Panax Notoginseng returnRt1; - } the intPush (intXinty) { + returnmerge (x, y); A } the intPop (int&x) { + intL =HEAP[X].L; - intR =HEAP[X].R; $Heap[l].par =l; $Heap[r].par =R; -HEAP[X].L = HEAP[X].R = Heap[x].dis =0; - returnmerge (L, R); the } - BOOLScan_d (int&num) {Wuyi Char inch;BOOLisn=false; the inch=GetChar (); - if(inch==eof)return false; Wu while(inch!='-'&& (inch<'0'||inch>'9'))inch=GetChar (); - if(inch=='-') {isn=true; num=0;} About Elsenum=inch-'0'; $ while(inch=getchar (),inch>='0'&&inch<='9'){ -num*=Ten, num+=inch-'0'; - } - if(IsN) num=-num; A return true; + } the intMain () { - while(Scan_d (N)) { $ for(inti =1; I <= N; ++i) { the Scan_d (heap[i].val); theHEAP[I].L = HEAP[I].R = Heap[i].dis =0; theHeap[i].par =i; the } - Scan_d (M); in intA, B, X, y; the while(M-- ) { the Scan_d (a); Scan_d (b); Aboutx =find (a); they =find (b); the if(x = =y) { thePuts ("-1" ); +}Else { -Heap[x].val/=2; the intPX =push (POP (x), x); BayiHeap[y].val/=2; the intPY =push (Pop (y), y); the -printf ("%d\n", heap[merge (px, py)].val); - } the } the } the return 0; the}
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Hdoj 1512 almost template left-leaning tree