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;}

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感觉这题的代码写的好挫。。。。