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POJ1270 Following Orders[拓扑排序所有方案 Kahn]

2024-08-13 00:55:15   218人阅读

Following Orders
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 4885 Accepted: 1973

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 ga b b fv w x y zv y x v z v w v

Sample Output

abfgabgfagbfgabfwxzvywzxvyxwzvyxzwvyzwxvyzxwvy

Source

Duke Internet Programming Contest 1993,uva 124
－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－
所有方案，需要回溯，用Kahn比较好
L← Empty list that will contain the sorted elementsS ← Set of all nodes with no incoming edgeswhile S is non-empty do    remove a node n from S    insert n into L    foreach node m with an edge e from nto m do        remove edge e from thegraph        ifm has no other incoming edges then            insert m into Sif graph has edges then    return error (graph has at least onecycle)else     return L (a topologically sortedorder)

就是找入度为0的点(最好用个stack,循环的话复杂的太高)，加入topo头部

感觉比dfs好，复杂度都是O(V+E)

 

本题回溯所有方案，复杂度乘上一个V；V很小，不用stack也可以；用个id比较方便吧

 

字符读入太坑人.........

 

//// main.cpp// poj1270//// Created by Candy on 9/11/16.//#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int N=30,M=55;char s[105];int a[N],num=0,n=0,id[N];int ch[N][N],topo[N],ind[N];void print(){    for(int i=1;i<=n;i++) printf("%c",(char)topo[i]+‘a‘-1);    printf("\n");}void dfs(int d){ //printf("dfs %d\n",d);    if(d==n+1){print();return;}    for(int i=1;i<=n;i++)        if(ind[i]==0){            ind[i]--; topo[d]=a[i];            for(int j=1;j<=ch[i][0];j++) ind[ch[i][j]]--;            dfs(d+1);            for(int j=1;j<=ch[i][0];j++) ind[ch[i][j]]++;            ind[i]++;        }}int main(int argc, const char * argv[]) {    while(fgets(s,105,stdin)){ //printf("p %s\n",s);        n=0;        memset(topo,0,sizeof(topo));        memset(ch,0,sizeof(ch));        memset(ind,0,sizeof(ind));        int len=strlen(s); //printf("len %d\n",len);        for(int i=0;i<len;i++)            if(s[i]>=‘a‘&&s[i]<=‘z‘) a[++n]=s[i]-‘a‘+1;        sort(a+1,a+1+n);        for(int i=1;i<=n;i++) id[a[i]]=i;                fgets(s,105,stdin);        len=strlen(s);        int last=0;        for(int i=0;i<=len;i++)            if(s[i]>=‘a‘&&s[i]<=‘z‘){                int t=s[i]-‘a‘+1;                t=id[t];                if(last==0) last=t;                else{ch[last][++ch[last][0]]=t;ind[t]++;last=0;}            }        dfs(1);        printf("\n");    }    return 0;}

 

 

 

 

POJ1270 Following Orders[拓扑排序所有方案 Kahn]


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