图的遍历（洛谷试炼场——提高历练地）

图的遍历

P2661 信息传递

//查找有向图最小环
#include<bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
#define ll long long
#define db double
#define PII pair<int, int>
#define pi acos(-1.0)
const int INF = 0x7fffffff;
const int N = 2e5 + 5;
const db eps = 1e-10;
using namespace std;
int cas, n, a[N], ru[N], chu[N], ans = INF;
bool vis[N];
int dfs(int now, int begin){  //遍历环
    if(a[now] == begin) return 1;
    chu[now]--;
    now = a[now];
    ru[now]--;
    if(!ru[now]) vis[now] = 0;
    return 1 + dfs(now, begin);
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin >> n;
    memset(ru, 0, sizeof(ru));
    memset(chu, 0, sizeof(chu));
    rep(i, 1, n){
        cin >> a[i];
        ru[a[i]]++, chu[i]++;  //所有点出度都为 1 
        vis[i] = 1;
    }
    rep(i, 1, n){  //将所有入度为 0 的点依次斩掉，最后只剩环
        if(!vis[i]) continue;
        int now = i;
        while(!ru[now]){
            chu[now]--, vis[now] = 0;
            now = a[now];
            ru[now]--;
        }
    }
    rep(i, 1, n){
        if(!vis[i]) continue;
        ans = min(ans, dfs(i, i));
    }
    cout << ans << endl;
}

P1330 封锁阳光大学

#include<bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
#define ll long long
#define PII pair<int, int>
const int N = 1e4 + 5;
using namespace std;
int m, n, vis[N], deep[N], flg = 1, a[N], ans = 0;
vector<int> edge[N];
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin >> n >> m;
    memset(a, 0, sizeof(a));
    rep(i, 1, m){
        int x, y; cin >> x >> y;
        edge[x].push_back(y), edge[y].push_back(x);
        a[x] = 1, a[y] = 1;
    }
    memset(vis, 0, sizeof(vis));
    rep(i, 1, n){
        if(vis[i]) continue;
        //BFS
        deep[i] = 1, vis[i] = 1;
        queue<int> record; record.push(i);
        queue<PII> q; q.push({i, 1});
        while(q.size()){
            int tp = q.front().first, cnt = q.front().second;
            q.pop();
            for(auto nxt : edge[tp]){
                if(vis[nxt]){
                    if(deep[nxt] &&/* abs(deep[nxt] - deep[tp]) % 2 == 0*/ deep[nxt] == deep[tp]){
                        puts("Impossible");
                        return 0;
                    }
                }
                else{
                    deep[nxt] = cnt + 1, vis[nxt] = 1;
                    q.push({nxt, deep[nxt]}), record.push(nxt);
                }
            }
        }
        int odd = 0, even = 0;
        while(record.size()){  //比较每棵树上的奇层和偶层数量的最小值
            int tp = record.front(); record.pop();
            if(deep[tp] % 2) even++;
            else odd++;
        }
        ans += min(odd, even);  //对于森林中每棵树累加最小值
    }
    cout << ans << endl;
}

P1341 无序字母对

//欧拉回路
#include<bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
#define all(x) x.begin(), x.end()
using namespace std;
int n, g[300][300];
map<char, int> mp;
vector<char> tmp, ans;
void In(char x){
    if(mp.count(x)) mp[x]++;
    else mp.insert({x, 1});
}
void dfs(char x){  //遍历欧拉回路
    rep(i, 65, 122){  //"A" = 65, "z" = 122
        if(g[x][i]){
            g[x][i] = g[i][x] = 0;
            dfs(i);
        }
    }
    ans.push_back(x);  //逆序存储
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin >> n;
    memset(g, 0, sizeof(g));
    rep(i, 1, n){
        char x[5]; cin >> x;
        In(x[0]), In(x[1]);
        g[x[0]][x[1]] = g[x[1]][x[0]] = 1;
    }
    for(auto now : mp)  //查找奇度数点
        if(now.second % 2) tmp.push_back(now.first);
    if(mp.size() > n + 1 || tmp.size() > 2 || tmp.size() == 1){
        puts("No Solution");
        return 0;
    }
    if(!tmp.size()){  //欧拉回路：所有点度数为偶数
        pair<char, int> head = *mp.begin();
        dfs(head.first);
    }
    else dfs(min(tmp[0], tmp[1]));  //欧拉路径：仅有两个奇度数点

    reverse(all(ans));
    rep(i, 0, ans.size() - 1) cout << ans[i];
}

P2921 Trick or Treat on the Farm G

#include<bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
const int N = 1e5 + 5;
using namespace std;
int n, nxt[N], dfn[N], ans[N], Begin;
bool vis[N];
int dfs(int now, int cnt){
    if(ans[now]) return cnt + ans[now] - 1;  //若从链走到某个环，直接返回
    if(vis[now]){  //遍历完环了
        Begin = now;  //记录环的起点
        return ans[now] = cnt - dfn[now];
    }
    vis[now] = 1;
    dfn[now] = cnt;  //记录时间戳
    dfs(nxt[now], cnt + 1);
    if(Begin){  //在环上
        if(now == Begin) Begin = 0;
        else ans[now] = ans[nxt[now]];
    }
    else ans[now] = ans[nxt[now]] + 1;
    vis[now] = 0;
    return ans[now];
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin >> n;
    rep(i, 1, n) cin >> nxt[i];
    rep(i, 1, n) cout << dfs(i, 1) << endl;
}