POJ1094 / ZOJ1060

#include <cstdio>
#include <cstring>
#include <stack>
#include <iostream>
using namespace std;
#define N 1005

int first[N] , in[N] , rec[N] , vis[N] , k;
char str[N][5];

struct Node
{
    int y , next;
}node[N<<1];

void add_edge(int x,int y)
{
    in[y]++;
    node[k].y = y , node[k].next = first[x];
    first[x] = k++;
}

int dag(int cnt , int n)
{
    int copy_in[N];
    for(int i = 0 ; i<n ; i++) copy_in[i] = in[i];
    int t = 0;
    stack<int> s;
    for(int i = 0 ; i<n ; i++){
        if(!copy_in[i] && vis[i])
        {
          //  cout<<"there: "<<i<<" "<<cnt<<" "<<n<<endl;
            t++;
            s.push(i);
        }
    }

    int flag = 1;
    for(int i = 0 ; i<cnt ; i++){
        if(s.empty()){
            return -1;
        }
        if(s.size() > 1) flag = 0;
        int u = s.top();
        s.pop();
        rec[i] = u;
        for(int i=first[u] ; i!=-1 ; i=node[i].next){
            int v = node[i].y;
            copy_in[v]--;
            if(!copy_in[v]) s.push(v);
        }
    }
    if(cnt == n && t == 1 && flag) return 1;
    else return 0;
}

int main()
{
  //  freopen("a.in" , "rb" , stdin);
    int n,m,a,b;
    while(~scanf("%d%d",&n,&m)){
        if(n==0 && m==0) break;
        memset(first , -1 , sizeof(first));
        memset(in , 0 , sizeof(in));
        memset(vis , 0 , sizeof(vis));

        for(int i = 0 ; i<m ; i++){
            scanf("%s",str[i]);
        }

        int cnt = 0;//计算当前传入的所有边中含有的点数
        k=0;
        for(int i=0 ; i<m ; i++){
            //add edge
            a = str[i][0] - 'A';
            if(!vis[a]){
                cnt++;
                vis[a]=1;
            }
            b = str[i][2] - 'A';
            if(!vis[b]){
                cnt++;
                vis[b]=1;
            }
          //  cout<<"here: "<<a<< " "<<b<<endl;
            add_edge(a,b);

            int falg = dag(cnt , n);
            if(falg == -1)
            {
                printf("Inconsistency found after %d relations.
" , i+1);
                break;
            }

            else if(falg == 1)
            {
                printf("Sorted sequence determined after %d relations: " , i+1);
                for(int i=0;i<n;i++)
                {
                    char c = 'A'+rec[i];
                    cout<<c;
                }
                printf(".
");
                break;
            }

            else{
                if(i<m-1) continue;
                printf("Sorted sequence cannot be determined.
");
            }
        }
    }
    return 0;
}

这道题就是逐步判断是否找到合适的顺序，找到了就停止将顺序输出

这里最大的问题是，在有向图中我们进行DAG查询的过程中，每一步都得保证只有一个点没有入度，否则多点没有入度，分不清先后顺序