CF858F Wizard's Tour

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麻了一道大水题因为我忘了图不一定联通调了 40 分钟...

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思路

一棵树的情况很好解决，按照拓扑序（也就是从叶子到根）进行匹配，且每次都最后匹配与父亲的连边（若边数为奇数则不匹配）

变成无向图后，考虑先走出一棵 DFS 树，然后和树的做法一模一样

其实这样的做法就是为了时刻保持原本的联通性

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代码

#include<iostream>
#include<fstream>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#define LL long long
#define FOR(i, x, y) for(int i = (x); i <= (y); i++)
#define ROF(i, x, y) for(int i = (x); i >= (y); i--)
#define PFOR(i, x) for(int i = he[x]; i; i = r[i].nxt)
inline int reads()
{
    int sign = 1, re = 0; char c = getchar();
    while(c < '0' || c > '9'){if(c == '-') sign = -1; c = getchar();}
    while('0' <= c && c <= '9'){re = re * 10 + (c - '0'); c = getchar();}
    return sign * re;
}
int n, m;
struct Node
{
    int to, nxt;
}r[400005]; int he[200005];
inline void Edge_add(int u, int v)
{
    static int cnt = 1;
    r[++cnt] = (Node){v, he[u]};
    he[u] = cnt;
}
int ans, p1[100005], p2[100005], p3[100005];
std::bitset<200005> vis, rvis, used;
int fa[200005];
void dfs(int now)
{
    vis[now] = 1;
    PFOR(i, now)
    {
        if(rvis[i >> 1]) continue; 
        int to = r[i].to;
        if(vis[to]) continue;
        rvis[i >> 1] = 1;
        fa[to] = i, dfs(to);
    }
    int line = 0;
    PFOR(i, now)
    {
        if(used[i >> 1]) continue;
        if((i ^ 1) == fa[now]) continue;
        if(!line) line = i;
        else
        {
            p2[++ans] = now, p1[ans] = r[line].to, p3[ans] = r[i].to,
            used[line >> 1] = used[i >> 1] = 1;
            line = 0;
        }
    }
    if(line && fa[now])
    {
        p2[++ans] = now, p1[ans] = r[line].to, p3[ans] = r[fa[now] ^ 1].to,
        used[line >> 1] = used[fa[now] >> 1] = 1;
        line = 0;
    }
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif
    n = reads(), m = reads();
    FOR(i, 1, m)
    {
        int u = reads(), v = reads();
        Edge_add(u, v), Edge_add(v, u);
    }
    FOR(i, 1, n) if(!vis[i]) dfs(i);
    printf("%d\n", ans);
    FOR(i, 1, ans) printf("%d %d %d\n", p1[i], p2[i], p3[i]);
    return 0;
}