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 M lines 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
传统的做法是先把输入的点加入并查集建立关系，然后开始判断询问，遇到destroy就将那两个点之间的关系断开，但是很明显，这样很难做到将其关系断开。。。
对啊，正常并查集做不到怎么办，不要一直被局限在正常思维，可以逆向思维想一下，那些destory的边我先不进行合并，将询问离线，倒着处理询问，这样就简单多了。。。
怎么找到最小标号的最大值呢，就要在合并的时候处理一下子了，只要每次优先权值小的指向权值大的，遇到权值相同的，则标号大的指向标号小的。。。。。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1);
const double eps = 1e-8;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
const int N = 10005;
int val[N];
int b[N];
int ans[50005];
pair<int,int>ve[20005];
map<pair<int,int>,int>mk;
typedef struct Node{
char s[15];
int x,y;
}Node;
Node node[50005];
int n,m,q;
void init()
{
for(int i = 0;i < n;++i){
b[i] = i;
}
}
int join(int x)
{
if(b[x] != x){
b[x] = join(b[x]);
}
return b[x];
}
void unin(int x,int y)
{
int tx = join(x);
int ty = join(y);
if(tx != ty){
if(val[tx] < val[ty]){
b[tx] = ty;
}else if(val[tx] > val[ty]){
b[ty] = tx;
}else if(tx < ty){
b[ty] = tx;
}else{
b[tx] = ty;
}
}
}
int main()
{
bool flag = false;
while(~scanf("%d",&n))
{
init();
if(flag){
printf("\n");
}else{
flag = true;
}
mk.clear();
for(int i = 0;i < n;++i){
scanf("%d",&val[i]);
}
scanf("%d",&m);
for(int i = 0;i < m;++i){
int x,y;
scanf("%d %d",&x,&y);
ve[i].fi = x;ve[i].se = y;
}
scanf("%d",&q);
for(int i = 0;i < q;++i){
scanf("%s",node[i].s);
if(node[i].s[0] == 'q'){
scanf("%d",&node[i].x);
}else{
scanf("%d %d",&node[i].x,&node[i].y);
mk[mp(node[i].x,node[i].y)] = 1;
mk[mp(node[i].y,node[i].x)] = 1;
}
}
for(int i = 0;i < m;++i){
if(mk[mp(ve[i].fi,ve[i].se)] == 0){
unin(ve[i].fi,ve[i].se);
}
}
memset(ans,inf,sizeof(ans));
for(int i = q - 1;i >= 0;--i){
if(node[i].s[0] == 'd'){
unin(node[i].x,node[i].y);
}else{
int tx = join(node[i].x);
if(val[tx] > val[node[i].x]){
ans[i] = tx;
}else{
ans[i] = -1;
}
}
}
for(int i = 0;i < q;++i){
if(ans[i] != inf){
printf("%d\n",ans[i]);
}
}
}
return 0;
}