POJ 1094 Sorting It All Out

Sorting It All Out

Time Limit: 1000ms

Memory Limit: 10000KB

This problem will be judged on PKU. Original ID: 1094
64-bit integer IO format: %lld      Java class name: Main

 

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.

Source

East Central North America 2001

 

解题：这题说的是边读入一些关系。如果当前出现矛盾的（有环），就记录出现矛盾的是第几组关系。如果当前组已经能够确定n个字母的关系，也要记录当前是第几组。如果当前关系不唯一。那么继续了。

 

判关系是唯一就是时刻判断队列中的元素是不是大于1.如果大于1，说明同时有几个点入度为0.排序就不确定了。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 105;
int n,m,cnt,in[maxn],tmp[maxn],result[maxn];
vector<int>g[maxn];
queue<int>q;
void init() {
    for(int i = 0; i < maxn; i++) {
        in[i] = 0;
        g[i].clear();
    }
}
bool Find(int x,int y) {
    for(int i = 0; i < g[x].size(); i++)
        if(g[x][i] == y) return false;
    return true;
}
int topSort() {
    while(!q.empty()) q.pop();
    for(int i = 0; i < n; i++)
        if(!in[i]) q.push(i);
    cnt = 0;
    bool uncertain = false;
    while(!q.empty()) {
        int u = q.front();
        if(q.size() > 1) uncertain = true;
        q.pop();
        result[cnt++] = u;
        for(int i = 0; i < g[u].size(); i++) {
            if(--in[g[u][i]] == 0) q.push(g[u][i]);
        }
    }
    if(cnt < n) return 1;
    if(uncertain) return 2;
    return 3;
}
int main() {
    bool ok;
    char u,v,op;
    int endPoint,flag;
    while(scanf("%d %d",&n,&m),n||m) {
        init();
        ok = false;
        getchar();
        flag = 2;
        for(int i = 1; i <= m; i++) {
            scanf("%c%c%c%*c",&u,&op,&v);
            if(ok) continue;
            if(op == '<' && Find(v-'A',u-'A')) {
                g[v-'A'].push_back(u-'A');
                in[u-'A']++;
            } else if(op == '>' && Find(u-'A',v-'A')) {
                g[u-'A'].push_back(v-'A');
                in[v-'A']++;
            }
            memcpy(tmp,in,sizeof(in));
            flag = topSort();
            if(flag != 2) {
                endPoint = i;
                ok = true;
            }
            memcpy(in,tmp,sizeof(in));
        }
        if(flag == 3) {
            printf("Sorted sequence determined after %d relations: ", endPoint);
            for(int i = cnt-1; i >= 0; i--)
                printf("%c",result[i]+'A');
            puts(".");
        } else if(flag == 1) printf("Inconsistency found after %d relations.\n",endPoint);
        else puts("Sorted sequence cannot be determined.");
    }
    return 0;
}