题意：给定森林，可以把两棵树连起来或者询问链上第k大。
解：启发式合并。
我一开始想到了启发式合并但是发现这样做之后一棵子树就不是一段连续的区间了，那就不能子树xxx了，很迷惘。
后来看了题解发现本来就不需要子树是连续区间......
每次把小的树暴力DFS重构fa[][]和重建主席树。
调了半天是因为lastans没有套上X[]......
注意并查集merge的时候可能有元素为0。无视之即可。
#include <cstdio>
#include <algorithm> const int N = , M = ; struct Edge {
int nex, v;
}edge[N << ]; int top; int X[N], e[N], n, val[N], temp, pw[N], fa[N][], tot, vis[N], father[N], siz[N], d[N], rt[N];
int sum[M], ls[M], rs[M];
char str[]; int find(int x) {
if(father[x] == x) {
return x;
}
return father[x] = find(father[x]);
} inline void merge(int x, int y) {
if(!x || !y) {
return;
}
x = find(x);
y = find(y);
if(x == y) {
return;
}
father[x] = y;
siz[y] += siz[x];
return;
} inline bool check(int x, int y) {
return find(x) == find(y);
} inline void add(int x, int y) {
top++;
edge[top].v = y;
edge[top].nex = e[x];
e[x] = top;
return;
} void add(int x, int &y, int p, int l, int r) {
if(!y || y == x) {
y = ++tot;
sum[y] = sum[x];
ls[y] = ls[x];
rs[y] = rs[x];
}
if(l == r) {
sum[y]++;
return;
}
int mid = (l + r) >> ;
if(p <= mid) {
add(ls[x], ls[y], p, l, mid);
}
else {
add(rs[x], rs[y], p, mid + , r);
}
sum[y] = sum[ls[y]] + sum[rs[y]];
return;
} void DFS(int x, int f) {
merge(x, f);
vis[x] = ;
fa[x][] = f;
d[x] = d[f] + ;
for(int j = ; j <= pw[n]; j++) {
fa[x][j] = fa[fa[x][j - ]][j - ];
}
rt[x] = ;
add(rt[f], rt[x], val[x], , temp);
for(int i = e[x]; i; i = edge[i].nex) {
int y = edge[i].v;
if(y == f) {
continue;
}
DFS(y, x);
}
return;
} inline void link(int x, int y) {
if(check(x, y)) {
printf("E1");
exit();
}
if(siz[find(x)] < siz[find(y)]) {
std::swap(x, y);
}
DFS(y, x);
add(x, y);
add(y, x);
return;
} inline int lca(int x, int y) {
if(d[x] > d[y]) {
std::swap(x, y);
}
int t = pw[n];
while(t >= && d[x] != d[y]) {
if(d[fa[y][t]] >= d[x]) {
y = fa[y][t];
}
t--;
}
if(x == y) {
return x;
}
t = pw[n];
while(t >= && fa[x][] != fa[y][]) {
if(fa[x][t] != fa[y][t]) {
x = fa[x][t];
y = fa[y][t];
}
t--;
}
return fa[x][];
} int Ask(int x, int y, int z, int w, int k, int l, int r) {
if(l == r) {
return r;
}
int mid = (l + r) >> , s = ;
s = sum[ls[x]] + sum[ls[y]] - sum[ls[z]] - sum[ls[w]];
if(k <= s) {
return Ask(ls[x], ls[y], ls[z], ls[w], k, l, mid);
}
else {
return Ask(rs[x], rs[y], rs[z], rs[w], k - s, mid + , r);
}
} inline int ask(int x, int y, int k) {
if(!check(x, y)) {
printf("E2");
exit();
}
int z = lca(x, y);
if(d[x] + d[y] - d[z] - d[z] + < k) {
printf("E3");
exit();
}
return Ask(rt[x], rt[y], rt[z], rt[fa[z][]], k, , temp);
} int main() { //freopen("in.in", "r", stdin);
//freopen("my.out", "w", stdout); int m, q;
scanf("%d", &n);
scanf("%d%d%d", &n, &m, &q);
for(int i = ; i <= n; i++) {
scanf("%d", &val[i]);
X[++temp] = val[i];
siz[i] = ;
father[i] = i;
}
for(int i = , x, y; i <= m; i++) {
scanf("%d%d", &x, &y);
add(x, y);
add(y, x);
}
for(int i = ; i <= n; i++) {
pw[i] = pw[i >> ] + ;
}
std::sort(X + , X + temp + );
temp = std::unique(X + , X + temp + ) - X - ;
for(int i = ; i <= n; i++) {
val[i] = std::lower_bound(X + , X + temp + , val[i]) - X;
}
for(int i = ; i <= n; i++) {
if(!vis[i]) {
DFS(i, );
}
}
/// build int lastans = ;
for(int i = , x, y, k; i <= q; i++) {
scanf("%s%d%d", str, &x, &y);
if(str[] == 'L') { // link
link(x ^ lastans, y ^ lastans);
}
else {
scanf("%d", &k);
lastans = X[ask(x ^ lastans, y ^ lastans, k ^ lastans)];
printf("%d\n", lastans);
}
} return ;
}
AC代码