"BZOJ-1455" Roman games can be stacked (left-leaning tree)

1455: Roman Games time limit:5 Sec Memory limit:64 MB
submit:1355 solved:561
[Submit] [Status] [Discuss] Description Roman Emperors liked to play killing games. There are n men in his army, and everyone is a separate regiment. A plane geometry test was recently held and everyone got a score. The emperor liked plane geometry, and he sniffed at those who scored very low. He decided to play such a game. It can send two kinds of commands: 1. Merger (I, J). Merge the regiment where I resides and the group where J is located into a regiment. If I, J has a person who is dead, then ignore the command. 2. Kill (i). Kill the person with the lowest score in the group where I was located. If I were dead, this command would be ignored. The emperor wanted him to publish a kill order, and the following generals reported the scores of the people who had been killed. (If this command is ignored, then 0 points) input the first line an integer n (1<=n<=1000000). n indicates the number of soldiers, and m represents the total number of commands. The second row n integers, where number I indicates the score of the soldier numbered I. (Fractions are integers between [0..10000]) The third line, an integer m (1<=m<=100000), 3+i Line describes the article I command. The command is in the following two forms: 1. M I J 2. K ioutput If the command is kill, the corresponding output is the score of the homicide. (If this person does not exist, output 0) Sample Input5
100 90 66) 99 10
7
M 1 5
K 1
K 1
M 2 3
M 3 4
K 5
K 4
Sample Output10
100
0
66
Hintsourcesolution

And pile up naked questions.

Three properties:

[Property 1] The key value of the node is less than or equal to the key value of its left and right child nodes.

[Property 2] The distance of the left child node of the node is not less than the distance of the right child node.

[Property 3] The left child node of the node is also a left-biased tree.

Code

#include <iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>using namespacestd;intRead () {intx=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9') {if(ch=='-') f=-1; Ch=GetChar ();}  while(ch>='0'&& ch<='9') {x=x*Ten+ch-'0'; Ch=GetChar ();} returnx*F;}#defineMAXN 1000010intd[maxn],son[maxn][2],a[maxn],n,fa[maxn],m;BOOLDEAD[MAXN];intFindintx) {    if(fa[x]==x)returnXElse returnfa[x]=find (Fa[x]);}intMergeintXinty) {    if(!x)returny; if(!y)returnx; if(a[x]>A[y]) swap (x, y); son[x][1]=merge (son[x][1],y); if(d[son[x][1]]>d[son[x][0]]) Swap (son[x][0],son[x][1]); D[X]=d[son[x][1]]+1; returnx;}intMain () {n=read ();  for(intI=1; i<=n; i++) a[i]=read ();  for(intI=1; i<=n; i++) fa[i]=i; M=read (); d[0]=-1;  for(intI=1; i<=m; i++)        {            Charopt[5];intx, y; scanf"%s", opt); if(opt[0]=='M') {x=read (), Y=read ();intFx=find (x), fy=Find (y), RT; if(dead[x]| | Dead[y])Continue; if(fx!=FY) RT=merge (fx,fy), fa[fx]=fa[fy]=RT; }            if(opt[0]=='K') {x=read ();intfx=find (x); if(Dead[x]) {Puts ("0");Continue;} DEAD[FX]=1; printf"%d\n", A[fx]); FA[FX]=merge (son[fx][0],son[fx][1]); FA[FA[FX]]=FA[FX]; }        }    return 0;}

"BZOJ-1455" Roman games can be stacked (left-leaning tree)