【BZOJ】4025: 二分图

题解

lct维护一个结束时间作为边权的最大生成树,每次出现奇环就找其中权值最小的那条边,删掉的同时还要把它标记上,直到这条边消失
如果有标记则输出No

边权通过建立虚点来维护

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 300005
//#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,T,cnt;
struct E_node {
    int u,v,s,t;
}E[MAXN];
vector<int> a[MAXN],b[MAXN];
bool vis[MAXN];
namespace lct {
    struct node {
	int lc,rc,fa,val,minq,siz;
	bool rev;
    }tr[MAXN];
#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
#define siz(u) tr[u].siz
    void Init() {
	val(0) = minq(0) = 0x7fffffff;
	for(int i = 1 ; i <= N ; ++i) {
	    val(i) = minq(i) = T + 2;
	    siz(i) = 1;
	}
    }
    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)));
	siz(u) = 1 + siz(lc(u)) + siz(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);
    }
    int Query_len(int u,int v) {
	Makeroot(u);Access(v);Splay(u);
	return siz(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::Query_len;
using lct::tr;
void Init() {
    read(N);read(M);read(T);
    lct::Init();
    for(int i = 1 ; i <= M ; ++i) {
	read(E[i].u);read(E[i].v);read(E[i].s);read(E[i].t);
	a[E[i].s + 1].pb(i);b[E[i].t + 1].pb(i);
	tr[i + N].siz = 1;tr[i + N].val = tr[i + N].minq = E[i].t;
    }
}
void Solve() {
    for(int i = 1 ; i <= T ; ++i) {
	int s = a[i].size();
	for(int j = 0 ; j < s ; ++j) {
	    int k = a[i][j];
	    if(E[k].u == E[k].v) {
		if(!vis[k]) {vis[k] = 1;++cnt;}
	    }
	    else if(!Connected(E[k].u,E[k].v)) {Link(k + N,E[k].u);Link(k + N,E[k].v);}
	    else {
		int t = Query(E[k].u,E[k].v);
		if(tr[t].val > tr[k + N].val) t = k + N;
		if((Query_len(E[k].u,E[k].v) / 2) % 2 == 0) {
		    if(!vis[t - N]) {vis[t - N] = 1;++cnt;}
		}
		if(t != k + N) {
		    Cut(t,E[t - N].u);Cut(t,E[t - N].v);
		    Link(k + N,E[k].u);Link(k + N,E[k].v);
		}
	    }
	}
	s = b[i].size();
	for(int j = 0 ; j < s ; ++j) {
	    int k = b[i][j];
	    Cut(E[k].u,k + N);Cut(E[k].v,k + N);
	    if(vis[k]) {vis[k] = 0;--cnt;}
	}
	if(cnt) puts("No");
	else puts("Yes");
    }
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
    return 0;
}
posted @ 2018-12-12 09:31  sigongzi  阅读(310)  评论(0编辑  收藏  举报