BZOJ 4229选择

BZOJ 4229选择

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题目分析：

看到删除，直接进行考虑倒过来进行加点，那么这样的话就就进行加边，然后查询两个点是否在同一个点边双中。虽然但是，这个东西真的可以用 LCT 进行维护，只需要加上一个并查集即可。那么并查集维护两个点是否属于同一个边双，如果是的话，那么利用一个并查集压缩这一条链。

总之就是分为两种情况。

1. 如果原来不相连的话，那么直接连上就行了
2. 如果在一颗 LCT 里面的话，那么直接把一个点提到根上然后直接缩这一条链即可。

Code:

//editor : DRYAYST
//Wo shi ge da SHA BI
#include<bits/stdc++.h>
#define g() getchar()
#define il inline
#define ull unsigned long long
#define eps 1e-10
#define ll long long
#define pa pair<int, int>
#define for_1(i, n) for(int i = 1; i <= (n); ++i)
#define for_0(i, n) for(int i = 0; i < (n); ++i)
#define for_xy(i, x, y) for(int i = (x); i <= (y); ++i)
#define for_yx(i, y, x) for(int i = (y); i >= (x); --i)
#define for_edge(i, x) for(int i = head[x]; i; i = nxt[i])
#define int long long
#define DB double
#define m_p make_pair
#define fi first
#define se second
using namespace std;
const int N = 1e6 + 10, INF = 0x7f7f7f7f, mod = 1e9 + 7;
il int qpow(int x, int k) {int ans = 1; while(k) {if(k & 1) ans = ans * x % mod; x = x * x % mod; k >>= 1; } return ans; }
il int Add(int x, int y) {return (x += y) %= mod;}
il int Del(int x, int y) {return (x = x - y + mod) % mod;}
il int Mul(int x, int y) {return x * y % mod;}
il int inv(int x) {return qpow(x, mod - 2); }
inline int re() {
    int x = 0, p = 1;
    char ch = getchar();
    while(ch > '9' || ch < '0') {if(ch == '-') p = -1; ch = getchar();}
    while(ch <= '9' and ch >= '0') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
    return x * p;
}
int n, m, q;
int ans[N], fa1[N], fa2[N];
int get1(int x) {return x == fa1[x] ? x : fa1[x] = get1(fa1[x]); }
int get2(int x) {return x == fa2[x] ? x : fa2[x] = get2(fa2[x]); }
set<pa> s; 
struct Node {int x, y, op; }Q[N], e[N];
#define ls c[x][0]
#define rs c[x][1]
int fa[N], c[N][2], st[N];
bool r[N];
il bool nroot(int x) {return c[get2(fa[x])][0] == x || c[get2(fa[x])][1] == x; }
il bool tp(int x) {return c[get2(fa[x])][1] == x; }
il void pushdown(int x) {if(r[x]){swap(c[ls][0], c[ls][1]); swap(c[rs][0], c[rs][1]); r[ls] ^= 1; r[rs] ^= 1; r[x] = 0; }}
il void rotate(int x) {
    int y = get2(fa[x]), z = get2(fa[y]), k = tp(x); 
    if(nroot(y)) c[z][tp(y)] = x; c[y][k] = c[x][!k]; c[x][!k] = y; 
    fa[c[y][k]] = y; fa[y] = x; fa[x] = z; 
}
il void splay(int x) {
    int y = x, z = 0; st[++z] = y;  while(nroot(y)) y = get2(fa[y]), st[++z] = y ;  while(z)  pushdown(st[z]), z--; 
    while(nroot(x)) { 
        int y = get2(fa[x]); 
        if(nroot(y)) {rotate(tp(x) == tp(y) ? y : x); } rotate(x); 
    }
}
il void access(int x) {int y = 0; for(; x; y = x, x = get2(fa[x])) splay(x), rs = y; }
il void makeroot(int x) {access(x); splay(x); swap(ls, rs), r[x] ^= 1; }
void BL(int x) { if(fa[x]) fa2[x] = get2(fa[x]); if(ls) BL(ls); if(rs) BL(rs); }
il void link(int x, int y) { 
    x = get2(x), y = get2(y); if(x == y) return; 
    if(get1(x) != get1(y)) {makeroot(x); fa[x] = y; fa1[get1(x)] = get1(y);}
    else {makeroot(x); access(y); splay(y); BL(y);}
}
#undef ls
#undef rs

signed main() {
    freopen("choice.in","r",stdin); freopen("choice.out","w",stdout); 
    n = re(), m = re(), q = re(); for_1(i, m) {int x = re(), y = re(); if(x > y) swap(x, y); e[i].x = x, e[i].y = y;}
    for_1(i, q) {char op; scanf("%s", &op); int x = re(), y = re(); if(x > y) swap(x, y); Q[i].x = x; Q[i].y = y;  Q[i].op = (op == 'Z') ? 1 : 2;  if(op == 'Z') s.insert(m_p(x, y));}
    for_1(i, n) fa1[i] = fa2[i] = i;  for_1(i, m) if(!s.count(m_p(e[i].x, e[i].y))) {link(e[i].x, e[i].y); }
    for(int i = q; i; --i) {if(Q[i].op == 1) link(Q[i].x , Q[i].y); else ans[i] = (get2(Q[i].x) == get2(Q[i].y)); }
    for_1(i, q) {if(Q[i].op == 2) printf(ans[i] ? "Yes\n" : "No\n");}
}
/*
7 8 7
1 2
1 3
1 4
2 3
3 4
3 7
7 4
5 6
Z 1 4
P 1 3
P 2 4
Z 1 3
P 1 3
Z 6 5
P 5 6
*/