Legal or Not
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5885 Accepted Submission(s):
2726
get together. It is so harmonious that just like a big family. Every day,many
"holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to
exchange their ideas. When someone has questions, many warm-hearted cows like
Lost will come to help. Then the one being helped will call Lost "master", and
Lost will have a nice "prentice". By and by, there are many pairs of "master and
prentice". But then problem occurs: there are too many masters and too many
prentices, how can we know whether it is legal or not?
We all know a
master can have many prentices and a prentice may have a lot of masters too,
it's legal. Nevertheless，some cows are not so honest, they hold illegal
relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the
same time, 3xian is HH's master,which is quite illegal! To avoid this,please
help us to judge whether their relationship is legal or not.
Please note
that the "master and prentice" relation is transitive. It means that if A is B's
master ans B is C's master, then A is C's master.
case, the first line contains two integers, N (members to be tested) and M
(relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each
contains a pair of (x, y) which means x is y's master and y is x's prentice. The
input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number
(0, 1, 2,..., N-1). We use their numbers instead of their names.
the messy relationship.
If it is legal, output "YES", otherwise "NO".
题意：输入数据n，m，表示有n个人接下来m行，每行输入x，y表示x是y的师父；
如果A是B的师父B是C的师父，则A是C的师父
如果A是B的师父，B又是A的师父则不合法输出No，如果合法输出YES
题解：1、如果输入的点中无不依赖定点的点（成环）输出no
2、最后结果中不依赖顶点的节点个数少于n不符合题意
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int n,m;
int map[110][110];
int vis[110];
void getmap()
{
int i,j,a,b;
memset(vis,0,sizeof(vis));
memset(map,0,sizeof(map));
for(i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
if(!map[a][b])
{
vis[b]++;
map[a][b]=1;
}
}
}
void tuopu()
{
int i,j,sum=0;
int ok=0;
queue<int>q;
while(!q.empty())
q.pop();
for(i=0;i<n;i++)
{
if(vis[i]==0)
{
sum++;
q.push(i);
}
}
if(sum==0) ok=1;//开始图中就不存在不依赖顶点的节点（成环）
else
{
int u,ans=0;
while(!q.empty())
{
u=q.front();
ans++;
q.pop();
for(i=0;i<n;i++)
{
if(map[u][i])
{
vis[i]--;
if(vis[i]==0)
q.push(i);
}
}
}
if(ans<n)//最后排序完成后不依赖顶点的节点个数小于n
ok=1;//即存在环不符合题意
}
if(ok==0)
printf("YES\n");
else
printf("NO\n");
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m),n|m)
{
getmap();
tuopu();
}
return 0;
}