In order to strengthen the defense ability, many stars in galaxy allied together and built many bidirectional tunnels to exchange messages. However, when the Galaxy War began, some tunnels were destroyed by the monsters from another dimension. Then many problems were raised when some of the stars wanted to seek help from the others.
In the galaxy, the stars are numbered from 0 to N-1 and their power was marked by a non-negative integer pi. When the star A wanted to seek help, it would send the message to the star with the largest power which was connected with star A directly or indirectly. In addition, this star should be more powerful than the star A. If there were more than one star which had the same largest power, then the one with the smallest serial number was chosen. And therefore, sometimes star A couldn't find such star for help.
Given the information of the war and the queries about some particular stars, for each query, please find out whether this star could seek another star for help and which star should be chosen.
Input
There are no more than 20 cases. Process to the end of file.
For each cases, the first line contains an integer N (1 <= N <= 10000), which is the number of stars. The second line contains N integers p0, p1, ... , pn-1 (0 <= pi <= 1000000000), representing the power of the i-th star. Then the third line is a single integer M (0 <= M <= 20000), that is the number of tunnels built before the war. Then Mlines follows. Each line has two integers a, b (0 <= a, b <= N - 1, a != b), which means star a and star b has a connection tunnel. It's guaranteed that each connection will only be described once.
In the (M + 2)-th line is an integer Q (0 <= Q <= 50000) which is the number of the information and queries. In the following Q lines, each line will be written in one of next two formats.
"destroy a b" - the connection between star a and star b was destroyed by the monsters. It's guaranteed that the connection between star a and star b was available before the monsters' attack.
"query a" - star a wanted to know which star it should turn to for help
There is a blank line between consecutive cases.
Output
For each query in the input, if there is no star that star a can turn to for help, then output "-1"; otherwise, output the serial number of the chosen star.
Print a blank line between consecutive cases.
Sample Input
2 10 20 1 0 1 5 query 0 query 1 destroy 0 1 query 0 query 1
Sample Output
1 -1 -1 -1
并查集，因为要删除边 所以普通的做法并不能解决，这里考虑离线的做法，把所有数据保存下来，第一次建立集合的时候不把 要删除的边加进去。然后从后往前查询，每当遇到标记为删除的边时，更新集合。
/* *********************************************** Author :PK28 Created Time :2015/8/17 8:57:25 File Name :4.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 0x3f3f3f3f #define maxn 50000+10 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; int val[20100]; bool cmp(int a,int b){ return a>b; } int n,q,m,fa[maxn],arr[25000][2],ans[60000]; vector<pair<int,int> >vp; map<pair<int,int>,int>mp; void init(){ for(int i=0;i<=n+5;i++)fa[i]=i; mp.clear(); vp.clear(); cle(ans); cle(arr); } int findfa(int x){ if(x==fa[x])return x; else return fa[x]=findfa(fa[x]); } void Union(int a,int b){ int x=findfa(a); int y=findfa(b); if(x==y)return ; if(val[x]>val[y])fa[y]=x; else if(val[x]<val[y])fa[x]=y; else{ //这里 的判断要注意 if(x<y) fa[y]=x; else fa[x]=y; } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int mark=0; while(cin>>n){ if(mark)printf(" "); init(); for(int i=0;i<n;i++)scanf("%d",&val[i]); scanf("%d",&m); for(int i=1;i<=m;i++){ scanf("%d%d",&arr[i][0],&arr[i][1]); if(arr[i][0]>arr[i][1])swap(arr[i][0],arr[i][1]); } scanf("%d",&q); char s[20]; int t,p; for(int i=1;i<=q;i++){ scanf("%s",s); if(s[0]=='q'){ scanf("%d",&t); vp.push_back(make_pair(t,-2)); } else{ scanf("%d%d",&t,&p); if(t>p)swap(t,p); mp[make_pair(t,p)]=1; vp.push_back(make_pair(t,p)); } } for(int i=1;i<=m;i++){ if(!mp[make_pair(arr[i][0],arr[i][1])]) Union(arr[i][0],arr[i][1]); } int j=0; for(int i=vp.size()-1;i>=0;i--){ int a=vp[i].first; int b=vp[i].second; if(b==-2){ int c=findfa(a); if(val[a]>=val[c])c=-1; ans[j++]=c; } else Union(a,b); } for(int i=j-1;i>=0;i--)printf("%d ",ans[i]); mark=1; } return 0; }