【LOJ】#121. 「离线可过」动态图连通性

题解

和BZOJ4025挺像的

就是维护边权是时间的最大生成树

删边直接删

两点未联通时直接相连，两点联通则找两点间边权小的一条边删除即可

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 500005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,M;
int id[MAXN],tot,pos[MAXN + 5005];
int op[MAXN],x[MAXN],y[MAXN];
int t[5005][5005];
namespace lct {
    struct node {
	int lc,rc,fa,val,minq;
	bool rev;
    }tr[MAXN * 2];
#define lc(u) tr[u].lc
#define rc(u) tr[u].rc
#define fa(u) tr[u].fa
#define val(u) tr[u].val
#define minq(u) tr[u].minq
#define rev(u) tr[u].rev
    void Init() {
	val(0) = minq(0) = 0x7fffffff;
	for(int i = 1 ; i <= N ; ++i) val(i) = minq(i) = M + 2;
    }
    void reverse(int u) {
	swap(lc(u),rc(u));
	rev(u) ^= 1;
    }
    void pushdown(int u) {
	if(rev(u)) {
	    reverse(lc(u));
	    reverse(rc(u));
	    rev(u) = 0;
	}
    }
    void update(int u) {
	minq(u) = val(u);
	minq(u) = min(minq(u),minq(lc(u)));
	minq(u) = min(minq(u),minq(rc(u)));
    }

    bool isRoot(int u) {
	if(!fa(u)) return true;
	else return rc(fa(u)) != u && lc(fa(u)) != u;
    }
    bool which(int u) {
	return rc(fa(u)) == u;
    }
    void rotate(int u) {
	int v = fa(u);
	if(!isRoot(v)) {(v == lc(fa(v)) ? lc(fa(v)) : rc(fa(v))) = u;}
	fa(u) = fa(v);fa(v) = u;
	if(u == lc(v)) {lc(v) = rc(u);fa(rc(u)) = v;rc(u) = v;}
	else {rc(v) = lc(u);fa(lc(u)) = v;lc(u) = v;}
	update(v);
    }
    void Splay(int u) {
	static int que[MAXN],qr;
	qr = 0;int x;
	for(x = u ; !isRoot(x) ; x = fa(x)) que[++qr] = x;
	que[++qr] = x;
	for(int i = qr ; i >= 1 ; --i) pushdown(que[i]);
	while(!isRoot(u)) {
	    if(!isRoot(fa(u))) {
		if(which(fa(u)) == which(u)) rotate(fa(u));
		else rotate(u);
	    }
	    rotate(u);
	}
	update(u);
    }
    void Access(int u) {
	for(int x = 0 ; u ; x = u , u = fa(u)) {
	    Splay(u);
	    rc(u) = x;
	    update(u);
	}
    }
    void Makeroot(int u) {
	Access(u);Splay(u);reverse(u);
    }
    void Link(int u,int v) {
	Makeroot(u);Makeroot(v);Splay(v);fa(v) = u;
    }
    void Cut(int u,int v) {
	Makeroot(u);Access(v);Splay(u);
	if(rc(u) == v) {rc(u) = 0;fa(v) = 0;update(u);}
    }
    int dfs(int u) {
	if(val(u) == minq(u)) return u;
	pushdown(u);
	if(minq(lc(u)) == minq(u)) return dfs(lc(u));
	else return dfs(rc(u));
    }
    int Query(int u,int v) {
	Makeroot(u);Access(v);Splay(u);
	return dfs(u);
    }
    bool Connected(int u,int v) {
	Makeroot(u);Access(v);Splay(u);
	int p = u;
	while(rc(p)) p = rc(p);
	if(p == v) return true;
	return false;
    }
}
using lct::Link;
using lct::Cut;
using lct::Makeroot;
using lct::Query;
using lct::Connected;
using lct::tr;
void Init() {
    read(N);read(M);
    tot = N;
    lct::Init();
    for(int i = 1 ; i <= M ; ++i) {
	read(op[i]);read(x[i]);read(y[i]);
	if(op[i] == 0) {
	    id[i] = ++tot;pos[tot] = i;
	    t[x[i]][y[i]] = t[y[i]][x[i]] = id[i];
	    tr[tot].val = tr[tot].minq = M + 1;
	}
	if(op[i] == 1) {
	    int k = t[x[i]][y[i]];
	    tr[k].val = tr[k].minq = i;
	    id[i] = k;
	}
    }
}
void Solve() {
    for(int i = 1 ; i <= M ; ++i) {
	if(op[i] == 0) {
	    if(!Connected(x[i],y[i])) {
		Link(x[i],id[i]);Link(id[i],y[i]);
	    }
	    else {
		int t = Query(x[i],y[i]);
		if(lct::tr[t].val < lct::tr[id[i]].val) {
		    Cut(t,x[pos[t]]);Cut(t,y[pos[t]]);
		    Link(id[i],x[i]);Link(id[i],y[i]);
		}
	    }
	}
	else if(op[i] == 1) {
	    Cut(x[i],id[i]);Cut(y[i],id[i]);
	}
	else {
	    if(Connected(x[i],y[i])) puts("Y");
	    else puts("N");
	}
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
    return 0;
}