poj1094 Sorting It All Out

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.
Source

East Central North America 2001

题意大概是给n个变量m个不等式，不等式之间有传递性，判断这些不等式是否矛盾，若不矛盾，判断能否求出每一对变量的关系，若能求出，判断最少利用前几个不等式。

做法：传递闭包+拓补排序。
利用floyd实现传递闭包，dp[i][j]=1表示i

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;
const int MAXN = 30;

int n,m,dp[MAXN][MAXN],d[MAXN];
bool flag;
char c[4];

inline int floyd() {
    memset(d,0,sizeof(d));
    for(register int k=1; k<=n; k++)
        for(register int x=1; x<=n; x++)
            for(register int j=1; j<=n; j++) {
                if(dp[x][k] && dp[k][j])
                    dp[x][j]=1;
            }
    for(register int i=1; i<=n; i++)
        for(register int j=i+1; j<=n; j++) {
            if(dp[i][i]==1) return 1;
            if(dp[i][j]==1) d[j]++;
            if(dp[j][i]==1) d[i]++;
            if(dp[i][j]==1 && dp[j][i]==1)
                return 1;
        }
    for(register int i=1; i<=n; i++)
        for(register int j=i+1; j<=n; j++)
            if(dp[i][j]==0 && dp[j][i]==0)
                return 2;
    return 3;
}

int main() {
    while(~scanf("%d%d",&n,&m)) {
        if(n==0 && m==0) break;
        flag=false;
        memset(dp,0,sizeof(dp));
        for(register int i=1; i<=m; i++) {
            scanf("%s",c+1);
            if(flag) continue;
            dp[c[1]-'A'+1][c[3]-'A'+1]=1;
            int u=floyd();
            if(u==1) {
                printf("Inconsistency found after %d relations.",i);
                flag=true;
                printf("
");
            } else if(u==3) {
                printf("Sorted sequence determined after %d relations: ",i);
                queue<int> q;
                for(register int i=1; i<=n; i++)
                    if(d[i]==0) {
                        q.push(i);
                        break;
                    }
                while(q.size()) {
                    int x=q.front();
                    q.pop();
                    printf("%c",x+'A'-1);
                    for(register int i=1; i<=n; i++)
                        if(dp[x][i]) {
                            d[i]--;
                            if(d[i]==0)
                                q.push(i);
                        }
                }
                printf(".
");
                flag=true;
            }

        }
        if(!flag)
            printf("Sorted sequence cannot be determined.
");
//      for(register int i=1; i<=n; i++) {
//          for(register int j=1; j<=n; j++) {
//              cout<<dp[i][j]<<" ";
//          }
//          cout<<endl;
//      }
    }
}