題目描述

Once 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.

一開始有n只孤獨的猴子，然后他們要打m次架，每次打架呢，都會拉上自己朋友最牛叉的出來跟別人打，打完之后戰斗力就會減半，每次打完架就會成為朋友（正所謂不打不相識o(∩_∩)o ）。問每次打完架之后那倆猴子最牛叉的朋友戰斗力還有多少，若朋友打架就輸出-1.

輸入輸出格式

輸入格式：

?

There 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.

有多組數據

?

輸出格式：

?

For each of the conflict, output -1 if the two monkeys know each other, otherwise output the strength value of the strongest monkey among all of its friends after the duel.

?

輸入輸出樣例

輸入樣例#1：?復制 5 20 16 10 10 4 5 2 3 3 4 3 5 4 5 1 5 輸出樣例#1：?復制 8 5 5 -1 10

說明

題目可能有多組數據

?

?

可并堆裸題

1 #include<cstdio> 2 #include<cmath> 3 #include<algorithm> 4 #include<iostream> 5 #include<queue> 6 using namespace std; 7 const int MAXN=200001; 8 #define ls T[x].ch[0] 9 #define rs T[x].ch[1] 10 int read() 11 { 12 int x=0,f=1;char ch=getchar(); 13 while(ch<'0' || ch>'9') {if(ch=='-')f=-1;ch=getchar();} 14 while(ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();} 15 return x*f; 16 } 17 int root,N,All; 18 struct node 19 { 20 int fa,dis,val,ch[2]; 21 }T[MAXN]; 22 int Merge(int x,int y) 23 { 24 if(!x) return y; 25 if(!y) return x; 26 if( T[x].val < T[y].val) swap(x,y); 27 rs=Merge(rs,y); 28 T[rs].fa=x; 29 if(T[ls].dis<T[rs].dis) swap(ls,rs); 30 T[x].dis=T[rs].dis+1; 31 return x; 32 } 33 int Find(int x) 34 { 35 while(T[x].fa) x=T[x].fa; 36 return x; 37 } 38 39 int main() 40 { 41 #ifdef WIN32 42 freopen("a.in","r",stdin); 43 #else 44 #endif 45 T[0].dis=-1; 46 while(scanf("%d",&N)!=EOF) 47 { 48 for(int i=1;i<=N;i++) T[i].val=read(),T[i].ch[0]=T[i].ch[1]=T[i].fa=T[i].dis=0; 49 int M=read(); 50 for(int i=1;i<=M;i++) 51 { 52 int x=read(),y=read(); 53 if(Find(x)==Find(y)) {printf("-1\n");continue;} 54 x=Find(x);y=Find(y); 55 int Max=T[x].val>T[y].val?x:y; 56 T[Max].val/=2; 57 printf("%d\n",T[Max].val); 58 T[ T[Max].ch[0] ].fa=T[ T[Max].ch[1] ].fa = 0; 59 int lson=T[Max].ch[0],rson=T[Max].ch[1]; 60 T[Max].ch[0]=T[Max].ch[1]=0; 61 Merge(Merge(lson,rson),Max); 62 Merge(Find(x),Find(y)); 63 } 64 } 65 }

?

總結

以上是生活随笔為你收集整理的洛谷P1456 Monkey King的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。