题目

P1347 排序

解析

打开一看拓扑排序，要判环。
三种情况

• 有环（存在矛盾）
• 没环但在拓扑排序时存在有两个及以上的点入度为0（关系无法确定）
• 除了上两种情况（关系可确定）

本来懒了一下，直接在排序时判环，然后一直WA，遂怒写tarjan判环，第一个点注意特判两个点相同的情况，注意重边。
然后就有了这又臭又长的

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1000;
int n, m, num, sum, tot;
int du[N], head[N], in[N], ans[N], dfn[N], low[N], size[N];
bool vis[N];
char s[N];

queue<int>q;
struct node {
    int v, nx;
} e[N];

template<class T>inline void read(T &x) {
    x = 0; int f = 0; char ch = getchar();
    while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    x = f ? -x : x ;
    return;
}

inline void add(int u, int v) {
    for (int i = head[u]; ~i; i = e[i].nx) if (e[i].v == v) return;
    e[++num].nx = head[u], e[num].v = v, head[u] = num, in[v]++;
}

int topo() {
    int tmp = 0, cnt = 0;
    for (int i = 1; i <= n; ++i) if (!du[i]) q.push(i), tmp++;
    if (tmp > 1) return 0;
    while (!q.empty()) {
        tmp = 0, cnt++;
        int u = q.front(); q.pop();
        ans[cnt] = u;
        for (int i = head[u]; ~i; i = e[i].nx) {
            int v = e[i].v;
            du[v]--;
            if (!du[v]) q.push(v), tmp++;
        }
        if (tmp > 1) return 0;
    }
    return 1;
}

stack<int>sta;
void tarjan(int u) {
    dfn[u] = low[u] = ++tot;
    vis[u] = 1;
    sta.push(u);
    for (int i = head[u]; ~i; i = e[i].nx) {
        int v = e[i].v;
        if (!dfn[v]) tarjan(v), low[u] = min(low[u], dfn[u]);
        else if (vis[v]) low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {
        int x = -1; sum++;
        while (x != u) {
            x = sta.top(), sta.pop();
            vis[u] = 0, size[sum]++;
        }
    }
}

int main() {
    memset(head, -1, sizeof head);
    read(n), read(m);
    for (int i = 1; i <= m; ++i) {
        scanf("%s", s);
        add(s[0] - 'A' + 1, s[2] - 'A' + 1);
        if(s[0] == s[2]) {
            printf("Inconsistency found after %d relations.", i);
            return 0;
        }
        sum = tot = 0;
        memset(dfn, 0, sizeof dfn);
        memset(low, 0, sizeof low);
        memset(vis, 0, sizeof vis);
        memset(size, 0, sizeof size);
        for (int j = 1; j <= n; ++j) if (!dfn[j]) tarjan(j);
        for (int j = 1; j <= sum; ++j) if (size[j] > 1) {
            printf("Inconsistency found after %d relations.", i);
            return 0;
        }
        memcpy(du, in, sizeof du);
        while (!q.empty()) q.pop();
        int k = topo();
        if (k == 1) {
            printf("Sorted sequence determined after %d relations: ", i);
            for (int j = 1; j <= n; ++j) printf("%c", ans[j] + 'A' - 1);
            printf(".");
            return 0;
        } else continue;
    }
    printf("Sorted sequence cannot be determined.");
    return 0;
}