Sicily 1949 && 1876 Basic Graph Problem(RMQ+并查集)

//RMQ+并查集,RMQ套模板即可，此题关键是并查集的使用
//sicily 1949 1876是同一道题，但是注意CASE的大小写，时间区别是5s和1s
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
int _max[100010][18],_min[100010][18];
bool ok;
int father[100010];
int n,query,st,ed,u,v;
void RMQ()//套模板即可
{
for(int j = 1;(1<<j) <= n;++j)
for(int i = 1;i + (1<<j) -1 <= n;++i)
{
_max[i][j] = max(_max[i][j-1],_max[i+(1<<(j-1))][j-1]);
_min[i][j] = min(_min[i][j-1],_min[i+(1<<(j-1))][j-1]);
}
}
int getfather(int x)//数据规模是10W，因此找爸爸这一步骤最好写成非递归形式
{
int temp = x;
while(x != father[x])
x = father[x];
father[temp] = x;//路径压缩
return x;
}
void connect(int x,int y)//如果X和Y的爸爸不同，就让他们拥有共同祖先
{
x = getfather(x);
y = getfather(y);
if(x != y)	father[y] = x;
}
int Max(int st,int ed)
{
int k = (int)(log((double)(ed - st + 1))/log(2.0));
return max(_max[st][k],_max[ed-(1<<k)+1][k]);
}
int Min(int st,int ed)
{
int k = (int)(log((double)(ed - st + 1))/log(2.0));
return min(_min[st][k],_min[ed-(1<<k)+1][k]);
}
int main()
{
//freopen("in.txt","r",stdin);
int caseN = 0;
bool first = 1;
while(scanf("%d",&n) != EOF)
{
if(!first)	printf("/n");
first = 0;
for(int i = 1;i <= n;++i)	father[i] = i;
printf("Case %d/n",++caseN);
for(int i = 1;i <= n;++i)
{
scanf("%d",&_max[i][0]);
_min[i][0] = _max[i][0];
}
RMQ();
scanf("%d",&query);
while(query--)
{
scanf("%d%d",&st,&ed);
u = Max(st,ed);
v = Min(st,ed);
connect(u,v);
}
scanf("%d",&query);
while(query--)
{
scanf("%d%d",&u,&v);
if(getfather(u) == getfather(v))	printf("YES/n");
else printf("NO/n");
}
}
return 0;
}