【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;
}