POJ 1094 Sorting It All Out【拓扑排序】

题目链接：

http://poj.org/problem?id=1094

题意：

给定前n个字母的大小关系，问你是否

1. 根据前xxx个关系得到上升序列
2. 所有关系都无法确定唯一的一个序列
3. 第xxx个关系导致出现环

分析：

此题坑略多。。。。

1. m大小没给！！这个很无语啊。。。数组开大点马上AC了。。。
2. 无法确定序列必须最后判断。
3. 一旦可以判断出上升序列，就不用管后面是否出现闭环了~~

where xxx is the number of relations processed at the time either a sorted sequence is determined or an inconsistency is found, whichever comes first, and yyy…y is the sorted, ascending sequence.

代码：

#include<iostream>
#include<queue>
#include<stack>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 1005;
vector<int>G[maxn];
int in[maxn];
int ef[maxn], et[maxn];
int n, m;
queue<int>s;
int vis[maxn];
int judge(int num)
{
     while(!s.empty()) s.pop();
     int cnt = 0;
     memset(in, 0,sizeof(in));
     memset(vis, 0, sizeof(vis));
     for(int i = 0; i < maxn; i++)
        G[i].clear();
    for(int i = 0; i <= num; i++){
        G[ef[i]].push_back(et[i]);
        in[et[i]]++;
        if(!vis[et[i]]) cnt++;
        if(!vis[ef[i]]) cnt++;
        vis[et[i]] = vis[ef[i]] = 1;
    }
    queue<int>q;
    for(int j =0; j < n; j++){
        if(vis[j] && !in[j])
                q.push(j);
    }
    int flg = 0;
    while(!q.empty()){
        int u = q.front();q.pop();
        if(q.size()) flg = 1;
        s.push(u);
        for(int k = 0; k < G[u].size(); k++){
            int v = G[u][k];
            in[v]--;
            if(!in[v]) q.push(v);
        }
    }
    //cout<<s.size()<<' '<<cnt<<endl;
    if(s.size() == n && !flg) return 3;
    return s.size() == cnt;
}
int main (void)
{
   while(scanf("%d%d",&n, &m) && n+m != 0){
    for(int i = 0; i <maxn; i++)
        G[i].clear();
    getchar();
    char a, b;
    for(int i = 0; i < m; i++){
        scanf("%c%*c%c", &a, &b);
        getchar();
        ef[i] = a - 'A';
        et[i] = b - 'A';
    }
    int i;
    int flag = 0;
    for(i = 0; i < m; i++){
        int t = judge(i);
        if(t == 3){flag = 1;break;}
        else if(t == 0){break;}
    }
    if(flag){
        cout<<"Sorted sequence determined after "<<i+1<<" relations: ";
        while(!s.empty()){cout<<(char)(s.front()+'A');s.pop();}
        cout<<'.'<<endl;
    } else if(i == m)  cout<<"Sorted sequence cannot be determined."<<endl;
    else cout<<"Inconsistency found after "<<i + 1<<" relations."<<endl;
   }
    return 0;
}

感觉这题的代码写的好挫。。。。