题面
Sol
首先每次加入边的两个点不联通，那么联通块的个数就要减(1)
那么考虑怎么做
莫名想到(LCT)
然后就不会了。。。
(orz)题解
维护一个每条边的数组，如果这个点加入后形成环，那么就把这个数组设为环内最先加入的边的编号，特判自环，然后替换这条边
没有替换为(0)
那么每次询问答案就是求([l, r])内该数组小于(l)的边的个数，答案为(n)-个数(自己(yy)一下就好)
主席树来求这个东西
代码很丑
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(4e5 + 5);
const int __(4e6 + 5);
IL ll Input(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, k, type, rpl[_], fa[_], cnt, id[_];
struct Edge{
int u, v;
} edge[_];
struct LCT{
# define ls ch[0][x]
# define rs ch[1][x]
int ch[2][_], fa[_], mn[_], val[_], S[_], rev[_];
IL LCT(){
Fill(val, 127); Fill(mn, 127);
}
IL bool Son(RG int x){
return ch[1][fa[x]] == x;
}
IL bool Isroot(RG int x){
return ch[0][fa[x]] != x && ch[1][fa[x]] != x;
}
IL void Update(RG int x){
mn[x] = min(val[x], min(mn[ls], mn[rs]));
}
IL void Reverse(RG int x){
if(!x) return;
swap(ls, rs); rev[x] ^= 1;
}
IL void Adjust(RG int x){
if(!rev[x]) return;
rev[x] = 0; Reverse(ls); Reverse(rs);
}
IL void Rotate(RG int x){
RG int y = fa[x], z = fa[y], c = Son(x);
if(!Isroot(y)) ch[Son(y)][z] = x; fa[x] = z;
ch[c][y] = ch[!c][x]; fa[ch[c][y]] = y;
ch[!c][x] = y; fa[y] = x;
Update(y);
}
IL void Splay(RG int x){
S[S[0] = 1] = x;
for(RG int y = x; !Isroot(y); y = fa[y]) S[++S[0]] = fa[y];
while(S[0]) Adjust(S[S[0]--]);
for(RG int y = fa[x]; !Isroot(x); Rotate(x), y = fa[x])
if(!Isroot(y)) Son(x) ^ Son(y) ? Rotate(x) : Rotate(y);
Update(x);
}
IL void Access(RG int x){
for(RG int y = 0; x; y = x, x = fa[x]) Splay(x), ch[1][x] = y, Update(x);
}
IL void Makeroot(RG int x){
Access(x); Splay(x); Reverse(x);
}
IL void Link(RG int x, RG int y){
Makeroot(x); fa[x] = y;
}
IL void Split(RG int x, RG int y){
Makeroot(x); Access(y); Splay(y);
}
IL void Cut(RG int x, RG int y){
Split(x, y); fa[x] = ch[0][y] = 0; Update(y);
}
# undef ls
# undef rs
} T;
struct Segment{
int ls[__], rs[__], size[__], rt[_], cnt;
IL void Modify(RG int &x, RG int l, RG int r, RG int v){
ls[++cnt] = ls[x]; rs[cnt] = rs[x]; size[cnt] = size[x] + 1; x = cnt;
if(l == r) return;
RG int mid = (l + r) >> 1;
if(v <= mid) Modify(ls[x], l, mid, v);
else Modify(rs[x], mid + 1, r, v);
}
IL int Query(RG int A, RG int B, RG int l, RG int r, RG int L, RG int R){
if(L <= l && R >= r) return size[A] - size[B];
RG int mid = (l + r) >> 1, ans = 0;
if(L <= mid) ans = Query(ls[A], ls[B], l, mid, L, R);
if(R > mid) ans += Query(rs[A], rs[B], mid + 1, r, L, R);
return ans;
}
} HJT;
IL int Find(RG int x){ return x == fa[x] ? x : fa[x] = Find(fa[x]); }
int main(RG int argc, RG char* argv[]){
cnt = n = Input(); m = Input(); k = Input(); type = Input();
for(RG int i = 1; i <= n; ++i) fa[i] = i;
for(RG int i = 1; i <= m; ++i){
edge[i].u = Input(); edge[i].v = Input();
if(edge[i].u == edge[i].v){
rpl[i] = i;
continue;
}
RG int fx = Find(edge[i].u), fy = Find(edge[i].v);
if(fx != fy){
T.val[++cnt] = i; id[i] = cnt;
T.Link(cnt, edge[i].u), T.Link(cnt, edge[i].v);
fa[fx] = fy;
}
else{
T.Split(edge[i].u, edge[i].v);
rpl[i] = T.mn[edge[i].v];
T.Cut(id[rpl[i]], edge[rpl[i]].u); T.Cut(id[rpl[i]], edge[rpl[i]].v);
T.val[++cnt] = i; id[i] = cnt;
T.Link(cnt, edge[i].u), T.Link(cnt, edge[i].v);
}
}
for(RG int i = 1; i <= m; ++i){
HJT.rt[i] = HJT.rt[i - 1];
HJT.Modify(HJT.rt[i], 0, m, rpl[i]);
}
for(RG int ans = 0, i = 1; i <= k; ++i){
RG int l = Input(), r = Input();
if(type) l ^= ans, r ^= ans;
printf("%d
", ans = n - HJT.Query(HJT.rt[r], HJT.rt[l - 1], 0, m, 0, l - 1));
}
return 0;
}