poj1270Following Orders

Following Orders

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3819		Accepted: 1514

Description

Order is an important concept in mathematics and in computer science. For example, Zorn's Lemma states: ``a partially ordered set in which every chain has an upper bound contains a maximal element.'' Order is also important in reasoning about the fix-point semantics of programs. 


This problem involves neither Zorn's Lemma nor fix-point semantics, but does involve order. 
Given a list of variable constraints of the form x < y, you are to write a program that prints all orderings of the variables that are consistent with the constraints. 


For example, given the constraints x < y and x < z there are two orderings of the variables x, y, and z that are consistent with these constraints: x y z and x z y. 

Input

The input consists of a sequence of constraint specifications. A specification consists of two lines: a list of variables on one line followed by a list of contraints on the next line. A constraint is given by a pair of variables, where x y indicates that x < y. 


All variables are single character, lower-case letters. There will be at least two variables, and no more than 20 variables in a specification. There will be at least one constraint, and no more than 50 constraints in a specification. There will be at least one, and no more than 300 orderings consistent with the contraints in a specification. 


Input is terminated by end-of-file. 

Output

For each constraint specification, all orderings consistent with the constraints should be printed. Orderings are printed in lexicographical (alphabetical) order, one per line. 


Output for different constraint specifications is separated by a blank line. 

Sample Input

a b f g
a b b f
v w x y z
v y x v z v w v

Sample Output

abfg
abgf
agbf
gabf

wxzvy
wzxvy
xwzvy
xzwvy
zwxvy
zxwvy

Source

题意：排序问题，给定两行字母，第一行为字母元素，第二行是字母间的先后关系，每个关系有一对字母组成，a<b.\

分析：拓补排序，先建图，求出每个字母的入度，再DFS每次取入度为一的节点并将该点子节点入度减一，重复dfs

最后说下此题有坑，输入中每组样例间可能有空行，每组样例中可能到处多空格，切记。

代码（有BUG，怪数据太水，竟然AC了，有空再写）：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<map>
#include<cmath>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
const int MAXM=1005;
int mmax,cur,ecnt,first[27],in[27],h[27],g[27][27];
int num[27],done[27],p[27];
int nex[MAXM],v[MAXM];
map<char,int>mp;
map<int,char>bmp;
int topsort(int n)
{
    queue<int>q;
    memset(h,0,sizeof h);
    memset(num,0,sizeof num);
    for(int i=1;i<=n;i++)
    {
        if(!in[i])q.push(i);
        if(!in[i]&&first[i]==-1)h[i]=-1;//自由因子层数计零
    }
    int cur=0;mmax=0;
    while(!q.empty())
    {
        int x=q.front();q.pop();
        num[h[x]+1]++;
        cur++;
        for(int e=first[x];~e;e=nex[e])
        {
            h[v[e]]=max(h[v[e]],h[x]+1);
            if(h[v[e]]>mmax)mmax=h[v[e]];//最高层数
            if(--in[v[e]]==0)q.push(v[e]);
        }
    }
    return n==cur;
}
void add_(int a,int b)
{
    v[ecnt]=b;
    nex[ecnt]=first[a];
    first[a]=ecnt++;
}
void dfs(int sum,int free)//sum为进队字母总数，free为进队自由因子数
{
    if(sum==cur)
    {
        for(int i=1;i<=cur;i++)
            putchar(bmp[p[i]]);
        printf("\n");
    }
    for(int i=1;i<=cur;i++)
    if(!done[i]){
        if(h[i]==-1)//自由因子不受层数限制，只要点到，就dfs
        {
            done[i]=1;
            p[sum+1]=i;
            dfs(sum+1,free+1);
            done[i]=0;
            continue;
        }
        if(sum-free<num[h[i]])continue;//已进队的非自由因子是否满足输出该层的要求
        done[i]=1;
        p[sum+1]=i;
        dfs(sum+1,free);
        done[i]=0;
    }
}
int main()
{
    int i,cas=0;char c,cc[1000];
    while(gets(cc))
    {
        if(!strlen(cc))continue;
        if(cas++>0)printf("\n");
        mp.clear();bmp.clear();
        int t=strlen(cc);
        vector<char>vec;
        for(i=0;i<t;i++)
            if(cc[i]>='a'&&cc[i]<='z')vec.push_back(cc[i]);
        t=vec.size();cur=0;
        sort(vec.begin(),vec.end());//将输入数据有序化，方便dfs
        for(i=0;i<t;i++)
            mp[vec[i]]=++cur,bmp[cur]=vec[i];

        memset(g,0,sizeof g);
        memset(in,0,sizeof in);
        memset(first,-1,sizeof first);
        int x,y,k=1;
        while((c=getchar())!='\n')
        {
            if(c==' ')continue;
            if(k==1){x=mp[c],k++;continue;}
            if(k==2)y=mp[c];k=1;
            if(!g[x][y]&&x&&y)
            {
                g[x][y]=1;
                add_(x,y);
                in[y]++;
            }
        }
        topsort(cur);
        num[0]=0;
        for(i=2;i<=mmax+1;i++)//num[]存的是输出该层的前提（即要想第3层输出就必须输完前两层），层数从零记起
            num[i]+=num[i-1];
        dfs(0,0);
    }
    return 0;
}