POJ--1094--Sorting It All Out【拓扑排序】

来源：互联网发布：阿里云域名前面是什么编辑：程序博客网时间：2024/04/30 05:47

链接：http://poj.org/problem?id=1094

题意&思路：直接拓扑排序。多解输出一串英文，有环输出一段英文，唯一解输出一段英文及排序结果。

细节：题目描述不是很清楚，如果不看discuss我肯定要WA出翔。

discuss里总结了两点关键的：

1. 输入一条边时如果此时拓扑有解就输出这个解，即使后面的边成有向环也不管了，所以每次输入的时候都得进行拓扑排序。

2. 判断存在有向环应先于判断多解。

这道题主要是题目坑爹。

#include<cstring>#include<string>#include<fstream>#include<iostream>#include<iomanip>#include<cstdio>#include<cctype>#include<algorithm>#include<queue>#include<map>#include<set>#include<vector>#include<stack>#include<ctime>#include<cstdlib>#include<functional>#include<cmath>using namespace std;#define PI acos(-1.0)#define MAXN 50100#define eps 1e-7#define INF 0x7FFFFFFF#define LLINF 0x7FFFFFFFFFFFFFFF#define seed 131#define MOD 1000000007#define ll long long#define ull unsigned ll#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1struct node{    int u,v,next;}edge[MAXN];int head[30],in[30],ans[30],IN[30];int n,m,cnt,sum;void add_edge(int a,int b){    edge[cnt].u = a;    edge[cnt].v = b;    edge[cnt].next = head[a];    head[a] = cnt++;}int toposort(){    int i;    for(i=0;i<n;i++){        in[i] = IN[i];    }    queue<int>q;    for(i=0;i<n;i++){        if(!in[i]) q.push(i);    }    sum = 0;    int flag = 0;    if(q.size()>1)  flag = 1;    while(!q.empty()){        int u = q.front();        q.pop();        ans[sum] = u;        sum++;        for(i=head[u];i!=-1;i=edge[i].next){            int next = edge[i].v;            in[next]--;            if(in[next] == 0)   q.push(next);        }        if(q.size()>1)  flag = 1;    }    if(sum!=n)  return 2;    else if(flag)   return 1;    return 3;}int main(){    int i,j;    char str[5];    while(scanf("%d%d",&n,&m),n||m){        memset(IN,0,sizeof(in));        memset(head,-1,sizeof(head));        int flag = 0;        cnt = 0;        int temp;        int p;        for(i=0;i<m;i++){            scanf("%s",str);//            cout<<str<<endl;            if(flag)    continue;            IN[str[2]-'A']++;            add_edge(str[0]-'A',str[2]-'A');            temp = toposort();            if(temp==3||temp==2){                flag = 1;                p = i + 1;            }        }        if(flag){            if(temp==3){                printf("Sorted sequence determined after %d relations: ",p);                    for(i=0;i<n;i++){                        printf("%c",ans[i]+'A');                    }                printf(".\n");            }            else if(temp==2){                printf("Inconsistency found after %d relations.\n",p);            }        }        else    printf("Sorted sequence cannot be determined.\n");    }    return 0;}

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