标签：log main algo ons 战斗力 nsis tput 最大 first
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.
不知道题意怎么办，那就百度翻译吧，。。。。。。。
算了我来说一下题解：这道题意思是猴子打架的话，就会找他们认识的猴子中最大的来打，然后战斗力减半，然后再合并，最终输出
总共合并后的树上，最大值（也就是堆顶）。
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<cmath> 5 #include<cstring> 6 using namespace std; 7 8 const int NN=1e5+7; 9 10 int n,m,a[NN]; 11 int r[NN],l[NN],d[NN],fa[NN]; 12 bool died[NN]; 13 14 int find(int num) 15 { 16 if (fa[num]!=num) return find(fa[num]); 17 else return num; 18 } 19 int merge(int x,int y) 20 { 21 if (!x) return y; 22 if (!y) return x; 23 if (a[x]<a[y]) swap(x,y); 24 r[x]=merge(r[x],y); 25 fa[r[x]]=x; 26 if (d[r[x]]>d[l[x]]) swap(r[x],l[x]); 27 d[x]=d[r[x]]+1; 28 return x; 29 } 30 int main() 31 { 32 //freopen("1.in","r",stdin); 33 //freopen("fzy.out","w",stdout); 34 d[0]=-1; 35 while (~scanf("%d",&n)) 36 { 37 memset(l,0,sizeof(r)); 38 memset(r,0,sizeof(r)); 39 for (int i=1;i<=n;i++) 40 { 41 fa[i]=i; 42 scanf("%d",&a[i]); 43 } 44 scanf("%d",&m); 45 for (int i=1;i<=m;i++) 46 { 47 int k,u,v; 48 scanf("%d%d",&u,&v); 49 int x=find(u),y=find(v); 50 if (x==y) printf("%d\n",-1); 51 else 52 { 53 fa[l[x]]=l[x],fa[r[x]]=r[x];54 fa[l[y]]=l[y],fa[r[y]]=r[y];//先各个独立 55 a[x]/=2,a[y]/=2; 56 int t1=merge(l[x],r[x]); 57 int t2=merge(l[y],r[y]); 58 fa[t1]=t1,fa[t2]=t2; 59 l[x]=l[y]=r[x]=r[y]=0;//拆开 60 t1=merge(t1,x); 61 t2=merge(t2,y); 62 fa[t1]=t1,fa[t2]=t2; 63 int t=merge(t1,t2);//再合并 64 fa[t]=t; 65 printf("%d\n",a[t]); 66 } 67 } 68 } 69 }
标签：log main algo ons 战斗力 nsis tput 最大 first
原文地址：http://www.cnblogs.com/fengzhiyuan/p/7471103.html
