【题解】CF#611 H-New Year and Forgotten Tree

　　有趣啊~手玩一下这棵树，发现因为连边只对相连点的位数有限制，我们可以认为是在往一棵已经有 m 个结点的树上挂叶子结点直到满足要求。（m = log(10) n）。注意由于 m 超级无敌小，我们可以直接爆搜初始树，然后 dinic 二分图匹配即可。（网络流：一边的点表示限制，另一边的点表示位数。每一条限制可以删去一个节点， 检验一下是否能够删完即可）。

#include <bits/stdc++.h>
using namespace std;
#define maxn 300000
#define INF 9999999
int n, m, cal[maxn], num[maxn], cur[maxn];
int tot, mark[7][7], rec[7][7], lev[maxn];
int S, T, Q[maxn], d[maxn], deg[maxn]; 
char s1[maxn], s2[maxn];
priority_queue <int, vector <int>, greater <int> > q;

int read()
{
    int x = 0, k = 1;
    char c; c = getchar();
    while(c < '0' || c > '9') { if(c == '-') k = -1; c = getchar(); }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * k;
}

struct edge
{
    int cnp, to[maxn], last[maxn], head[maxn], f[maxn], F[maxn];
    edge() { cnp = 2; }
    void add(int u, int v, int fl)
    {
    //    cout << "*****" << u << " " << v << " " << fl << endl;
        to[cnp] = v, f[cnp] = F[cnp] = fl, last[cnp] = head[u], head[u] = cnp ++;
        to[cnp] = u, f[cnp] = F[cnp] = 0, last[cnp] = head[v], head[v] = cnp ++;
    }
}E1;

struct node
{
    int u, v;
}id[maxn];

bool bfs()
{
    queue <int> q; 
    memset(lev, 0, sizeof(lev)); lev[S] = 1; 
    q.push(S);
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        for(int i = E1.head[u]; i; i = E1.last[i])
        {
            int v = E1.to[i];
            if(E1.f[i] && !lev[v])
            {
                lev[v] = lev[u] + 1;
                q.push(v);
            }
        }
        if(lev[T]) return 1;
    }
    return 0;
}

int dfs(int u, int nf)
{
    if(u == T || !nf) return nf;
    int tf = 0;
    for(int i = E1.head[u]; i; i = E1.last[i])
    {
        int v = E1.to[i]; 
        if(!E1.f[i] || lev[v] != lev[u] + 1) continue;
        int af = dfs(v, min(E1.f[i], nf));
        tf += af, nf -= af;
        E1.f[i] -= af, E1.f[i ^ 1] += af;
    }
    return tf;
}

int dinic()
{
    int nf = 0;
    while(bfs()) nf += dfs(S, INF);
    return nf;
}

void dfs2(int u)
{
    for(int i = E1.head[u]; i; i = E1.last[i])
    {
        int v = E1.to[i]; if(!E1.f[i ^ 1]) continue;
        if(u > m && u <= tot + m && v >= 1 && v <= m) 
            for(int j = 1; j <= E1.f[i ^ 1]; j ++)
            {
                int t = u - m; t = (id[t].u == v) ? id[t].v : id[t].u;
                printf("%d %d
", num[t], cur[v] ++);
            }
        if(u == S) dfs2(v);
    }
}

void Search(int now)
{
    if(now >= m - 1)
    {
        int t = 1;
        for(int i = 1; i <= m; i ++) d[i] = deg[i];
        for(int i = 1; i <= m; i ++) if(!d[i]) q.push(i);
        memset(mark, 0, sizeof(mark));
        while(!q.empty() && t <= m - 2)
        {
            int u = q.top(); q.pop();
            int x = u, y = Q[t]; if(x > y) swap(x, y);
            mark[x][y] = 1; d[Q[t]] --;
            if(!d[Q[t]]) q.push(Q[t]); t ++;
        }
        if(q.size() >= 2)
        {
            int x = q.top(); q.pop(); int y = q.top(); q.pop();
            mark[x][y] = 1;
        }else if(q.size() >= 1) q.pop();
        for(int i = 2; i < E1.cnp; i ++) E1.f[i] = E1.F[i];
        for(int i = E1.head[S]; i; i = E1.last[i])
        {
            int v = E1.to[i]; if(!v) continue; v -= m;
            int x = id[v].u, y = id[v].v;
            if(x > y) swap(x, y);
            E1.f[i] -= mark[x][y]; 
            if(E1.f[i] < 0) return;
        }
        if(dinic() == n - m) 
        {
            for(int i = 1; i <= m; i ++)
                for(int j = i + 1; j <= m; j ++)
                    if(mark[i][j]) printf("%d %d
", num[i], num[j]);
            for(int i = 1; i <= m; i ++) cur[i] ++;
            dfs2(S);
            exit(0);
        }
        return;
    }
    for(int i = 1; i <= m; i ++) 
        deg[i] ++, Q[now] = i, Search(now + 1), deg[i] --;
}

int main()
{
    n = read(); int t = n;
    while(t) { m ++; t /= 10; }
    for(int i = 1, l = 1; i <= m; l *= 10, i ++) num[i] = cur[i] = l;
    for(int i = 1, l = 1; i < m; l *= 10, i ++) cal[i] = l * 10 - num[i]; 
    cal[m] = n - num[m] + 1; S = 0, T = m * m + m + 3;
    for(int i = 1; i < n; i ++)
    {
        scanf("%s%s", s1 + 1, s2 + 1);
        int l1 = strlen(s1 + 1), l2 = strlen(s2 + 1);
        if(l1 > l2) swap(l1, l2); rec[l1][l2] ++;
    }
    for(int i = 1; i <= m; i ++)
        for(int j = i; j <= m; j ++)
        {
            id[++ tot].u = i, id[tot].v = j;
            E1.add(S, tot + m, rec[i][j]);
            E1.add(tot + m, i, INF); E1.add(tot + m, j, INF);
        }
    for(int i = 1; i <= m; i ++) E1.add(i, T, cal[i] - 1);
    Search(1);
    printf("-1
");
    return 0;
}