洛谷-1347

洛谷-1347

思路

此题解的思路再加上这篇blog的代码实现。

注意：本题要求的不是一个拓扑排序就可以了，实际上是要求一条链的拓扑排序。

Code

#include <bits/stdc++.h>
using namespace std;
#define _u_u_ ios::sync_with_stdio(false), cin.tie(nullptr)
#define cf int _o_o_;cin>>_o_o_;for (int Case = 1; Case <= _o_o_;Case++)
#define SZ(x) (int)(x.size())
inline void _A_A_();
signed main() {_A_A_();return 0;}

using ll = long long;
// #define int long long
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
const int N = 210, M = 5010;
const int inf = 0x3f3f3f3f;

struct edge {
    int to, nxt;
}e[610];

int vis[26];    // 0->未访问；1->正在访问；-1->已经访问过
int idx, h[610], n , m, now=1;

void add(int u,int v) {
    e[idx].to = v;
    e[idx].nxt = h[u];
    h[u] = idx++;
}

vector<int> res;    // 记录拓扑序
set<int> st;    // 记录现在已经出现过的点（实际上没啥用

bool dfs(int u) {
    vis[u] = 1;
    for (int i = h[u];i != -1;i =e[i].nxt) {
        if (vis[e[i].to] == 1) return 1;
        if (vis[e[i].to] == 0 && dfs(e[i].to)) return 1;
    }
    vis[u] = -1;
    res.push_back(u);
    return 0;
}

void topo() {
    res.clear();
    memset(vis, 0,sizeof vis);
    for (auto i : st) { // 实际上可以从0到n-1
        if (!vis[i] && dfs(i)) {
            cout << "Inconsistency found after " << now << " relations.\n";
            exit(0);
        }
    }
    if (SZ(res) == n) {
        int cnt = 0;    // 这里用来判断是否是链
        for (int i = n - 1;i >= 1;i -- ) {  // 判断每两个相邻的节点是否连边
            bool flag = 0;
            int a = res[i], b = res[i - 1]; // 数据很水，这里写成a = i, b = i-1可以拿90分。
            for (int k = h[a];k != -1;k = e[k].nxt) {
                if (e[k].to == b) {
                    flag = 1;
                }
            }
            if (flag) cnt++;
        }
        if (cnt == n - 1) { // 是链
            cout << "Sorted sequence determined after " << now << " relations: ";
            for (int i = n - 1;i >= 0;i--) {
                char t = res[i] + 'A';
                cout << t ;
            }
            cout  << ".\n";
            exit(0);
        }
    }
}

inline void _A_A_() {
    #ifdef LOCAL
    freopen("in.in", "r", stdin);
    #endif
    _u_u_;
    memset(h, -1, sizeof h);
    cin >> n >> m;
    string s;
    for (;now <= m;now++)  {
        cin >> s;
        add(s[0] - 'A', s[2] - 'A');
        st.insert(s[0] - 'A');
        st.insert(s[2] - 'A');
        topo();
    }
    cout << "Sorted sequence cannot be determined.\n";
}