POJ 1094 Sorting It All Out(拓扑排序)

题目链接；
POJ 1094 Sorting It All Out
题意：
给出$n$，代表$n$个元素，和$m$条关系，问在哪条关系时可以确定顺序或者出现冲突或者不能确定顺序。
数据范围：$2 \leq n \leq 26$
分析；
主要是不能确定顺序的判断。
当队列中多余一个元素时是不能确定顺序的，但是这时有可能是冲突的。所以要还要判断是否冲突。
对于每条新加关系，进行一遍拓扑排序判断就好了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#include <queue>
#include <vector>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N = 100;
const int MAX_M = 600;

int degree[MAX_N], ans[MAX_N];
vector<int> vec[MAX_N];
queue<int> que;
char s[10];
pair<int, int> edge[MAX_M];

int TopoSort(int n)
{
    while(!que.empty()) que.pop();
    for(int i = 0; i < n; ++i) {
        if(degree[i] == 0) que.push(i);
    }
    int total = 0, flag = 0;
    while(!que.empty()) {
        if(que.size() > 1) flag = 1; // cannnot be determined
        int cur = que.front();
        que.pop();
        ans[total++] = cur;
        for(int i = 0; i < vec[cur].size(); ++i) {
            int to = vec[cur][i];
            degree[to]--;
            if(degree[to] == 0) {
                que.push(to);
            }
        }
    }
    if(flag == 0 && total == n) return 1; //determined 
    for(int i = 0; i < n; ++i) {
        if(degree[i] > 0) return -1; //no solution
    }
    return 0;
}

int main()
{
    int n, m;
    while(~scanf("%d%d", &n, &m) && (n || m)) {
        for(int i = 0; i < m; ++i) {
            scanf("%s", s);
            edge[i].first = s[0] - 'A', edge[i].second = s[2] - 'A';
        }
        int flag = 0, id;
        for(int i = 1; i <= m; ++i) {
            memset(degree, 0, sizeof(degree));
            for(int j = 0; j < n; ++j) { vec[j].clear(); }
            for(int j = 0; j < i; ++j) {
                int u = edge[j].first, v = edge[j].second;
                degree[v]++;
                vec[u].push_back(v);
            }
            flag = TopoSort(n);
            if(flag == 1 || flag == -1) {
                id = i;
                break;
            } 
        }
        if(flag == 0) printf("Sorted sequence cannot be determined.\n");
        else if(flag == -1) printf("Inconsistency found after %d relations.\n", id);
        else {
            printf("Sorted sequence determined after %d relations: ", id);
            for(int i = 0; i < n; ++i) {
                printf("%c", ans[i] + 'A');
            }
            printf(".\n");
        }
    }   
    return 0;
}