POJ 1094 拓扑排序

Description:
      规定对于一个只有大写字母的字符串是有大小顺序的。如ABCD.即A<B。B<C。C<D。那么问题来了。现在第一行给你n， m代表序列里只会出现前n的大写字母、以后的第2到m行给你m组。有三个字符。即“第一个字符” “<”"第二个字符"...请你给出答案。在第几组之后确定了排列顺序或者第几组之后出现矛盾。还是直到输入结束也没有排列顺序也没有出现矛盾。

思路很简单、就是把前n个字符转换成0~n-1的数字。然后，，每次输入都要拓扑一次。如果出现了固定顺序或者矛盾。以后就可以只输有输入操作了。如果直到最后这两种情况都没有出现的话。就是无法确定。

附代码：WA了好久。。仍然有地方不懂。。但是是人家自己敲得啦，，，~~~~(>_<)~~~~

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;

int map[27][27];
int degree[27];
int top[27];
int tot;
int n, m;
queue<int>q;

void init()
{
    tot = 0;
    memset(map, 0, sizeof(map));
    memset(degree, 0, sizeof(degree));
    memset(top, 0, sizeof(top));
}

int topsort()
{
    int l = 0;
    int flag = 0;
    while(!q.empty())
        q.pop();
    int dd[30];
    memcpy(dd, degree, sizeof(degree));
    for (int i=0; i<n; ++i)
    {
        if (dd[i] == 0)
        {
            q.push(i);
        }
    }
    while(!q.empty())
    {
        if (q.size() > 1)
           flag = 1;
        int k = q.front();
        q.pop();
        top[l++] = k;
        for (int i=0; i<n; ++i)
        {
            if (map[k][i] == 1)
               {
                   if (--dd[i] == 0)
                    q.push(i);
               }
        }
    }
    if (l != n)
        return 2;
    else if (flag == 1)
        return 1;
    // T_T 为什么必须先判断是不是有矛盾。再判断是不是无法确定呢。也就是。有时候同时有两个入度为0的点。并不能说明就是无法判断。
    //所以不能在flag == 1处。改为直接return 1.....
    
//      if (flag == 1)
//        return 1;
//      else if (l != n)
//        return 2;
    return 0;
}
int main()
{
    char a, b, temp;
    int i, j;
    int aa, bb;
    int mao, que;
    int now = 0;
    while(cin >> n >> m)
    {
        init();
        if (n == 0 && m == 0)
            break;
        mao = 0;
        que = 0;
        now = 0;
        for (i=0; i<m; ++i)
        {
            cin >> a >> temp >> b;
            now++;
            if (!mao && !que)
            {
                if (map[b-‘A‘][a-‘A‘] == 1)
                {
                    mao = 1;
                    cout << "Inconsistency found after " << now <<  " relations.\n";
                    continue;
                }
                if (map[a-‘A‘][b-‘A‘] == 0)
                {
                    map[a-‘A‘][b-‘A‘] = 1;
                    degree[b-‘A‘]++;
                }
               int res = topsort();
                if (res == 0)
                {
                    cout << "Sorted sequence determined after " << now << " relations: ";
                    for (int j=0; j<n; ++j)
                    {
                        cout << char(top[j] + ‘A‘);
                    }
                    cout << ".\n";
                    que = 1;
                }
                else if (res == 2)
                {
                    cout << "Inconsistency found after " << now <<  " relations.\n";
                    mao = 1;
                }
            }
        }
        if (!mao && !que)
        {
            cout << "Sorted sequence cannot be determined.\n";
        }
    }
    return 0;
}