hdu 1512 左偏樹

2019-11-11 05:54:19

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題目：

Monkey King

Time Limit: 10000/5000 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5571 Accepted Submission(s): 2395PRoblem DescriptionOnce in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its own way and none of them knows each other. But monkeys can't avoid quarrelling, and it only happens between two monkeys who does not know each other. And when it happens, both the two monkeys will invite the strongest friend of them, and duel. Of course, after the duel, the two monkeys and all of there friends knows each other, and the quarrel above will no longer happens between these monkeys even if they have ever conflicted.Assume that every money has a strongness value, which will be reduced to only half of the original after a duel(that is, 10 will be reduced to 5 and 5 will be reduced to 2).And we also assume that every monkey knows himself. That is, when he is the strongest one in all of his friends, he himself will go to duel. InputThere are several test cases, and each case consists of two parts.First part: The first line contains an integer N(N<=100,000), which indicates the number of monkeys. And then N lines follows. There is one number on each line, indicating the strongness value of ith monkey(<=32768).Second part: The first line contains an integer M(M<=100,000), which indicates there are M conflicts happened. And then M lines follows, each line of which contains two integers x and y, indicating that there is a conflict between the Xth monkey and Yth. OutputFor each of the conflict, output -1 if the two monkeys know each other, otherwise output the strongness value of the strongest monkey in all friends of them after the duel. Sample Input

520161010452 33 43 54 51 5 Sample Output855-110
分析：
每一只猴子看做一個結點，建立左偏樹，樹根維護最大值。若兩結點樹根相同，說明兩只猴子是朋友，輸出-1；
否則，兩根節點值減半，刪除，重新與原樹合并得兩新樹，兩新樹合并后返回根節點值。
本題僅涉及到左偏樹部分操作。
左偏樹入門：
http://wenku.baidu.com/view/1c18eff9aef8941ea76e051b.html
http://wenku.baidu.com/view/004aa1ee19e8b8f67c1cb9d4.html?from=search
http://sunmoon-template.blogspot.ca/2014_12_01_archive.html
代碼：
#include<iostream>#include<cstring>#include<cstdio>#include<cstdlib>#include<ctype.h>    //tower()#include<set>#include<iomanip>// cout<<setprecision(1)<<fixed<<a;#include<vector>   #include<time.h>  #include<assert.h>  //assert#include<cmath>	#include<algorithm>#include<bitset>#include<limits.h>#include<stack>#include<queue>using namespace std;const int maxn=100050;const int inf=INT_MAX-100;struct node{    int val,dis,lt,rt,fa;//fa 父節點}ltree[maxn];int n,q,t;void maketree(int a,int k){//結點成樹 a結點值為k     ltree[a].val=k;    ltree[a].fa=a;    ltree[a].lt=ltree[a].rt=0;    ltree[a].dis=(a==0)?-1:0;}int find(int x){//非遞歸路徑壓縮，返回所在樹根節點     int s,k,j;    s=k=x;    while(s!=ltree[s].fa) s=ltree[s].fa;    while(k!=s){        j=ltree[k].fa;        ltree[k].fa=s;        k=j;    }    return s;}int merge(int a,int b){//兩棵樹合并 改變rt,dis,fa     if(a==0) return b;    if(b==0) return a;    if(ltree[a].val<ltree[b].val) swap(a,b);    ltree[a].rt=merge(ltree[a].rt,b);//b接在a最右邊 	int &l=ltree[a].lt,&r=ltree[a].rt;//左右子樹的引用，取別名，不另占內存，值同時修改        ltree[r].fa=a;//ltree[l].fa=    if(ltree[l].dis<ltree[r].dis) swap(l,r);//維持左偏性     if(r==0) ltree[a].dis=0;	else ltree[a].dis=ltree[r].dis+1;    return a;}int solve(int a,int b){	int tmp,ta,tb;	a=find(a);	b=find(b);	if(a==b) return -1;//	ltree[ltree[a].lt].fa=ltree[a].lt;//	ltree[ltree[a].rt].fa=ltree[a].rt;	ltree[a].val>>=1;	tmp=merge(ltree[a].lt,ltree[a].rt);	ltree[a].dis=ltree[a].lt=ltree[a].rt=0;    ta=merge(tmp,a);//	ltree[ltree[b].lt].fa=ltree[b].lt;//	ltree[ltree[b].rt].fa=ltree[b].rt;	ltree[b].val>>=1; 	tmp=merge(ltree[b].lt,ltree[b].rt);	ltree[b].dis=ltree[b].lt=ltree[b].rt=0;    tb=merge(tmp,b);        tmp=merge(ta,tb);    ltree[tmp].fa=ltree[ta].fa=ltree[tb].fa=tmp;    return ltree[tmp].val;}int main(){//795MS	3540K    while(~scanf("%d",&n)){	    for(int i=1;i<=n;++i){	        scanf("%d",&t);	        maketree(i,t);	    }//	    maketree(0,0);	    scanf("%d",&q);	    int c,d;	    for(int i=0;i<q;++i){	        scanf("%d%d",&c,&d);	        printf("%d/n",solve(c,d));	    }    }    return 0;    }