nyoj 349 Sorting It All Out（拓扑排序度的理解）

Sorting It All Out

时间限制：3000 ms | 内存限制：65535 KB
难度：3

描述
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.

输入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.

输出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.

样例输入

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

样例输出

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

来源POJ
上传者
陈玉
输入n和m，n表示26个字母前n个字母，m表示有多少个关系，然后输入m个关系，判断是否这n个字母存在一个排序关系

如果存在输出在几个关系之后就输出几个关系之后就可以确定，比如第一个测试数据，前四个关系输入之后，就输出结果

后两个关系输入不用管，，如果存在环那么就输出冲突，如果不能确定次序就输出不能确定。

拓扑排序的理解，度的判断

#include<stdio.h>
#include<string.h>
#define M 30
int re[M],path[M][M],du[M],d[M];
char str[8];
int topo(char s[],int n){
int i,j,k;
int A=s[0]-'A';
int B=s[2]-'A';
if(path[A][B]==0){
path[A][B]=1;
d[B]++;//因为每一次调用函数的时候要保持原来的图的数据，所以用d数组保存度不变
}
for(i=0;i<n;i++)
du[i]=d[i];//du数组来判断当前函数度的变化
int flag=1,cnt,pos=0;
for(i=0;i<n;i++){
cnt=0;
for(j=0;j<n;j++){
if(du[j]==0){
cnt++;
k=j;
}
}
if(cnt==0) return -1;//cnt等于0表示存在环，即存在冲突
else if(cnt>1)	flag=0;//存在多个度为0的点，即次序不能确定
du[k]--;
re[pos++]=k;
for(j=0;j<n;j++){
if(path[k][j]==1)
du[j]--;
}
}
if(flag) return 1;
return 0;
}
void result(int n,int m){
int i,j;
int ok=1;
for(i=0;i<m;i++){
scanf("%s",str);
if(ok){
int t=topo(str,n);
if(t==1){
printf("Sorted sequence determined after %d relations: ",i+1);
for(j=0;j<n;j++)
printf("%c",char(re[j]+'A'));
printf(".\n");
ok=0;
}
else if(t==-1){
printf("Inconsistency found after %d relations.\n",i+1);
ok=0;
}
}
}
if(ok)	printf("Sorted sequence cannot be determined.\n");
}
int  main(){
int n,m;
char str[8];
while(scanf("%d%d",&n,&m),n+m){
if(m<n-1){//关系不够不能确定次序
for(int i=0;i<m;i++)
scanf("%s",str);
printf("Sorted sequence cannot be determined.\n");
}else{
memset(d,0,sizeof(d));
memset(path,0,sizeof(path));
result(n,m);
}
}
return 0;
}