HDU 3342 Legal or Not（强连通）

Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6071    Accepted Submission(s): 2820

Problem Description

ACM-DIY is a large QQ group where many excellent acmers 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.

 

Input

The input consists of several test cases. For each 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.

 

Output

For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".

 

Sample Input

3 20 11 22 20 11 00 0

 

Sample Output

YESNO

 

题意 ：一个可以有多个师傅，或者多个徒弟。 师傅的师傅的一样是自己的师傅，徒弟的徒弟的，一样是自己的徒弟。

         但是一个人不能即是一个人的师傅，也是这个人的徒弟。问 如果存在这种混乱关系的，就输出NO,  否则输出 YES.

题解：强连通判断各个点是否各自为一个强连通分量，即强连通分量为n个为YES，否则为NO。

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <string>#include <map>#include <cstdlib>#include <cmath>#include <vector>#include <set>#include <queue>using namespace std;const int maxn=1e2+5;struct Edge {    int to,nxt;} edge[maxn];int head[maxn],tot;int low[maxn],dfn[maxn],Stack[maxn];int id,top;int scc;int n,m;bool Instack[maxn];void addedge(int u,int v) {    edge[tot].to=v;    edge[tot].nxt=head[u];    head[u]=tot++;}void tarjan(int u) {    int v;    low[u]=dfn[u]=++id;    Stack[top++]=u;    Instack[u]=true;    for(int i=head[u]; i!=-1; i=edge[i].nxt) {        v=edge[i].to;        if(!dfn[v]) {            tarjan(v);            if(low[v]<low[u])                low[u]=low[v];        } else if(Instack[v]&&low[u]>dfn[v])            low[u]=dfn[v];    }    if(low[u]==dfn[u]) {        scc++;        do {            v=Stack[--top];            Instack[v]=false;        } while(v!=u);    }}void solve() {    id=scc=top=0;    memset(Instack,false,sizeof(Instack));    memset(dfn,false,sizeof(dfn));    for(int i=0; i<n; i++) {        if(!dfn[i])            tarjan(i);    }    if(scc==n)        puts("YES");    else        puts("NO");}int main() {    while(~scanf("%d%d",&n,&m)) {        if(n==0)            break;        tot=0;        memset(head,-1,sizeof(head));        while(m--) {            int u,v;            scanf("%d%d",&u,&v);            addedge(u,v);        }        solve();    }    return 0;}