传递闭包:sorting it all out

给定n个不等式 判断是否有逻辑错误

floyd传递闭包问题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int N=1010;
int read()
{
	int x=0,f=0,c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	return f?-x:x;
}

struct Edge
{
	int to,next;
}e[N*N];
int head[N],cnt,deg[N];
void add(int a,int b){e[++cnt]=(Edge){b,head[a]};head[a]=cnt;deg[b]++;}
int n,m;
bool d[N][N];
queue<int> q;
int ret;
void clean()
{
	for(int i=1;i<=n;i++)
	{
		head[i]=deg[i]=0;
		for(int j=1;j<=n;j++) d[i][j]=0;
	}
	ret=cnt=0;
	while(q.size()) q.pop();
}

int floyd()
{
	for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				d[i][j] |= d[i][k]&d[k][j];		
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if( d[i][j] &&d[j][i]) return -1;			
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		{
			if(i==j) continue;
			else if(!d[i][j]&&!d[j][i]) return 0;
		}	
	return 1;
}

void topo()
{
	
	for(int i=1;i<=n;i++)
		if(!deg[i])  q.push(i);
	while(q.size())
	{
		int x=q.front(); q.pop();
		putchar(x+'A'-1);
		for(int i=head[x];i;i=e[i].next)
		{
			int y=e[i].to;
			deg[y]--;
			if(!deg[y]) q.push(y);
		}
	}
	puts(".");
}

int main()
{
	loop:
	while( ( n=read() )&& ( m=read() ) )
	{
		clean();
		for(int i=1;i<=m;i++)
		{
			int x=getchar()-'A'+1;	bool opt=getchar()=='>';	int y=getchar()-'A'+1;
			getchar();
			if(opt) d[x][y]=1,add(y,x);
			else d[y][x]=1,add(x,y);
			ret=floyd();
			if(ret==1)
			{
				printf("Sorted sequence determined after %d relations: ", i);
				topo();
				goto loop;
			}
			if(ret==-1)  
			{
				printf("Inconsistency found after %d relations.\n",i); 
				goto loop;
			}
		}
		if(ret==0) puts("Sorted sequence cannot be determined."); 
	}
	return 0;
}

这个题目的易错点在于:
1 矛盾的优先级高于推不出来结果 因此要优先判断
2. 关于需不需要判断i和j相等的问题 矛盾和推不出结果是两种情况
3. 传递闭包的关系输出: 拓扑排序

标签：闭包,sorting,sequence,int,ret,include,out,getchar
来源： https://www.cnblogs.com/juruoHBr/p/15779397.html