POJ1094拓扑排序问题

这道题WA好多次才通过，中间有很多没考虑到，加上一点小失误，不过算是把题解出来了。

这道题的思想是拓扑排序问题，需要考虑的是拓扑排序的结果只能是一种，多种就不能把各个字母的大小比较出来，每次输入一条变就拓扑排序一次。

附上代码和测试实例

View Code

#include "stdio.h"
#include "string.h"
#define M 100
int g[M][M];
int q[M],p[M];
bool useif[M],h[M];
int top=0,n,m;
int cr = 0;
int tp[M];
int q1[M];
int fun()
{
    int i;
    bool use = false;
    cr = 0;
    top=0;
    memset(useif,0,sizeof(useif));
    for(i = 1;i<=n;i++)
    {
        q[i] = q1[i];
    }
    for (i=1;i<=n;i++)
    {
        if(h[i] && q[i] == 0)
        {
            useif[i] = true;
            p[top++] = i;
        }
    }
    if(top>1)
        use = true;
    while(top>0)
    {
        int x = p[--top];
        tp[cr++] = x;
        for(i=1;i<=n;i++)
        {
            if(!useif[i] && g[x][i]==1)
            {
                q[i]--;
                if(q[i] == 0)
                {
                    useif[i] = true;
                    p[top++] = i;
                }
                
            }
        }
        if(top > 1)
            use = true;
    }
    for (i=1;i<=n;i++)
    {
        if(q[i])
            return -1;
    }
    if(!use && cr == n)
        return 1;

    return 0;
}
int main()
{
    char s[4];
    int a,b,i,rs;
    int x,y;
    bool falg;
    while(scanf("%d %d",&n,&m)!=EOF &&  (n!=0 || m!=0) )
    {
        falg = false;
        memset(g,0,sizeof(g));
        memset(q,0,sizeof(q));
        memset(p,0,sizeof(p));
        memset(h,0,sizeof(h));
        memset(q1,0,sizeof(q1));
        for(i=1;i<=m;i++)
        {
            scanf("%s",s);    
            a = s[0] - 'A' +1;
            b = s[2] - 'A' +1;
            h[a] = true;
            h[b] = true;
            if(g[a][b]==0) q1[b]++;
            g[a][b] = 1;    
            if(!falg)
                rs = fun();
                
            if(rs == -1 && !falg)
            {
                x = i;
                falg = true;
            }
            else if(rs == 1 && !falg)
            {
                y = i;
                falg = true;
            }
        }
        if(rs == -1)
            printf("Inconsistency found after %d relations.\n",x);
        else if(rs == 1)
        {
            printf("Sorted sequence determined after %d relations: ",y);
            for(i=0;i<n;i++)
                printf("%c",tp[i]+64);
            printf(".\n");

        }
        else
            printf("Sorted sequence cannot be determined.\n");
    }
    return 0;
}

输入数据：

View Code

4 6
D<B
D<A
D<C
C<A
C<B
A<B
10 40
C<I
E<G
A<J
F<B
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C<B
E<H
G<I
D<B
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A<B
J<I
D<G
A<E
C<G
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H<G
C<A
F<J
B<G
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C<F
B<J
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D<A
F<I
A<H
C<E
F<H
A<G
B<I
F<A
H<J
F<G
F<E
C<J
15 60
B<M
M<H
N<G
N<E
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B<D
A<J
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15 60
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26 325
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5 10
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20 158
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10 40
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H<B
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20 70
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O<B
5 10
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B<E
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A<B
A<D
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B<C
10 20
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F<H
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E<H
C<J
C<H
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F<G
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6 11
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B<F
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A<F
C<E
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9 23
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H<A
C<E
I<H
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20 34
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D<O
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B<N
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20 190
B<E
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N<S
N<R
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M<J
B<J
F<B
G<B
R<O
Q<T
Q<G
D<G
M<F
C<T
H<E
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Q<F
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H<O
H<S
H<I
P<S
F<E
F<T
M<O
A<O
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H<N
M<K
Q<N
P<E
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P<O
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C<D
D<O
A<S
C<B
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M<G
R<J
N<O
L<I
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B<O
J<E
C<J
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Q<A
Q<S
L<B
K<T
O<T
A<G
H<G
C<R
P<B
H<Q
C<E
A<E
I<J
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N<T
C<O
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N<A
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N<G
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A<K
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A<F
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P<G
R<E
D<I
F<K
I<S
G<J
H<F
G<S
L<F
P<I
M<L
G<I
N<P
B<T
L<P
J<K
C<H
B<K
A<I
G<E
G<T
N<M
M<B
H<R
D<J
M<E
F<G
Q<M
D<S
H<B
N<J
Q<E
G<O
C<N
H<A
P<T
Q<B
L<S
R<S
N<L
L<G
Q<R
F<O
A<R
L<E
Q<P
E<T
M<R
M<S
L<O
A<L
R<P
H<J
L<K
M<P
A<M
L<T
R<T
L<R
S<J
N<E
F<I
A<D
D<F
Q<O
J<O
O<E
C<L
A<B
N<K
N<F
G<K
D<T
N<I
H<L
M<I
L<J
0 0

输出结果：

Sorted sequence determined after 6 relations: DCAB.
Sorted sequence determined after 29 relations: CFDAEBHGJI.
Sorted sequence cannot be determined.
Inconsistency found after 48 relations.
Sorted sequence determined after 318 relations: GMNVKCHFOBRJDZLPXAYSEIWQTU.
Sorted sequence determined after 25 relations: ABCDEFGHIJKLMNOPQRSTUVWXYZ.
Sorted sequence determined after 7 relations: BADCE.
Sorted sequence determined after 158 relations: HNAMGIPCTBJFRLEODKSQ.
Inconsistency found after 35 relations.
Sorted sequence cannot be determined.
Sorted sequence determined after 7 relations: ABDEC.
Sorted sequence cannot be determined.
Sorted sequence determined after 11 relations: ACDBFE.
Sorted sequence cannot be determined.
Sorted sequence cannot be determined.
Sorted sequence determined after 179 relations: CHQNAMDLRFPGISBJOKET.

转载于:https://www.cnblogs.com/asen32/archive/2012/07/09/2583722.html