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;}