• POJ1094 Sorting It All Out(拓扑排序)


    Sorting It All Out
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 30110   Accepted: 10411

    Description

    An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D. in this problem, we will give you a set of relations of the form A < B and ask you to determine whether a sorted order has been specified or not.

    Input

    Input consists of multiple problem instances. Each instance starts with a line containing two positive integers n and m. the first value indicated the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of the uppercase alphabet. The second value m indicates the number of relations of the form A < B which will be given in this problem instance. Next will be m lines, each containing one such relation consisting of three characters: an uppercase letter, the character "<" and a second uppercase letter. No letter will be outside the range of the first n letters of the alphabet. Values of n = m = 0 indicate end of input.

    Output

    For each problem instance, output consists of one line. This line should be one of the following three:

    Sorted sequence determined after xxx relations: yyy...y.
    Sorted sequence cannot be determined.
    Inconsistency found after xxx relations.

    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.

    Sample Input

    4 6
    A<B
    A<C
    B<C
    C<D
    B<D
    A<B
    3 2
    A<B
    B<A
    26 1
    A<Z
    0 0
    

    Sample Output

    Sorted sequence determined after 4 relations: ABCD.
    Inconsistency found after 2 relations.
    Sorted sequence cannot be determined.

    题意:这题题意好蛋疼啊!!!一个一个给边,一个一个的判断,先判断是否有环。然后再判断是否确定序列。最后前两者都不满足,则此序列不确定。
    收获:1.知道只有有环时才k<n。2.入度为0的点不唯一,则此序列不确定。3.入度很重要。
    下面是能AC代码,但是 0 0会有问题,不知道为什么?还请各位指教。
    #include <cstdio>
    #include <iostream>
    //#include <cstdlib>
    #include <algorithm>
    #include <ctime>
    #include <cmath>
    #include <string>
    #include <cstring>
    #include <stack>
    #include <queue>
    #include <list>
    #include <vector>
    #include <map>
    #include <set>
    using namespace std;
    
    const int INF=0x3f3f3f3f;
    const double eps=1e-10;
    const double PI=acos(-1.0);
    
    const int maxn=5000;
    struct Edge
    {
        int u, v, next;
    };
    Edge edge[maxn];
    int head[maxn];
    int num, n,m;
    void init_edge()
    {
        num = 0;
        memset(head, -1, sizeof(head));
    }
    void addedge(int u, int v)
    {
        edge[num].u = u;
        edge[num].v = v;
        edge[num].next = head[u];
        head[u] = num++;
    }
    int ans;
    int topo[maxn];
    int in[maxn];
    int topsort()
    {
        queue<int> q;
        int indeg[26];
        for(int i = 65; i < 65+n; i++)
        {
            indeg[i] = in[i];
            if(indeg[i] == 0) q.push(i);
        }
        int k = 0;
        int flag=0;
        while(!q.empty())
        {
            if(q.size()>1)
                flag = 1;
            int u = q.front();
            q.pop();
            topo[k++] = u;
            for(int i = head[u]; i != -1; i = edge[i].next)
            {
                int v = edge[i].v;
                indeg[v]--;
                if(indeg[v]==0)
                    q.push(v);
            }
        }
        if(k < n)
            return 0;
        if(flag)
            return -1;
        else
        return 1;
    }
    char s[5];
    int main()
    {
        while(~scanf("%d%d", &n, &m))
        {
            if(n==0 && m==0)
               break;
            init_edge();
            memset(in, 0, sizeof(in));
            int flag2 = 0;
            int flag3 = 0;
            for(int j = 1; j <= m; j++)
            {
                scanf("%s",s);
                if(!flag2&&!flag3)
                {
                    in[s[2]]++;
                    addedge(s[0],s[2]);
                    int res = topsort();
                    if(res == 0)
                    {
                        printf("Inconsistency found after %d relations.
    ", j);
                        flag2 = 1;
                    }
                    if(res == 1)
                    {
                        printf("Sorted sequence determined after %d relations: ", j);
                        for(int i=0; i < n; i++)
                            printf("%c", topo[i]);
                        printf(".
    ");
                        flag3 = 1;
                    }
                }
            }
            if(!flag2&&!flag3)
                puts("Sorted sequence cannot be determined.");
    
        //puts("QAQ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ZP-Better/p/4706138.html
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