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【POJ】1094 Sorting It All Out（拓扑排序）

2024-07-25 06:47:25   221人阅读

http://poj.org/problem?id=1094

原来拓扑序可以这样做，原来一直sb的用白书上说的dfs。。。。。。。。。。。。

拓扑序只要每次将入度为0的点加入栈，然后每次拓展维护入度即可。。

我是个大sb，这种水题调了一早上。。

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>using namespace std;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << (#x) << " = " << (x) << endl#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }#define printarr1(a, b) for1(_, 1, b) cout << a[_] << ‘\t‘; cout << endlinline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }inline const int max(const int &a, const int &b) { return a>b?a:b; }inline const int min(const int &a, const int &b) { return a<b?a:b; }const int N=27, oo=~0u>>1;int n, m, e[N][N], in[N], s[N], vis[N], top, tp[N], ans[N], tot;int toposort() {	top=tot=0; int num=0, ret=0;	for1(i, 1, n) if(vis[i] && in[i]==0) s[++top]=i;	for1(i, 1, n) if(vis[i]) ++num;	for1(i, 1, n) tp[i]=in[i];	while(top) {		if(top>1) ret=-1;		int u=s[top--]; ans[tot++]=u;		for1(i, 1, n) if(e[u][i] && --tp[i]==0) s[++top]=i;	}	if(tot!=num) return ret=1;	return ret;}int main() {	while(~scanf("%d%d", &n, &m) && !(n==0 && m==0)) {		CC(e, 0); CC(in, 0); CC(vis, 0);		int u, v; char s[10];		int flag=0;		for1(i, 1, m) {			scanf("%s", s);			if(flag==1) continue;			u=s[0]-‘A‘+1; v=s[2]-‘A‘+1; 			e[u][v]=vis[u]=vis[v]=1; ++in[v];			flag=toposort(); 			if(flag==0) {				if(tot!=n) continue;				printf("Sorted sequence determined after %d relations: ", i);				rep(j, tot) printf("%c", ‘A‘+ans[j]-1);				puts(".");				flag=1;			}			else if(flag==1) printf("Inconsistency found after %d relations.\n", i);		}		if(flag==-1) puts("Sorted sequence cannot be determined.");	}	return 0;}

 

 

 

 

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 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 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

【POJ】1094 Sorting It All Out（拓扑排序）


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