In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn’t know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?
Input
The first line contains a single integer T, the number of test cases. And followed T cases.
The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly.
Output
The output should contain T lines. Write ‘Yes’ if the cave has the property stated above, or ‘No’ otherwise.
Sample Input
1
3 3
1 2
2 3
3 1
Sample Output
Yes
题意 给一个有向图,问是否对于任意的两个点,都可以相互到达。
是不是和强连通很像,其实我少说了一个很重要的条件,就是这个任意的两个点:没有规定那个是起始点,那个是终点,都可以当起点或者终点。所以和强连通还是很不一样。 其实这样的图叫做弱连通图;
对于一个DAG(有向图求scc+缩点后的新图(有向无环图))图来说,其实这样题意的图(弱连通图),就是只要有一个起点,它可以走过所有的点,并到达终点。 这样的图符合题意(弱连通图);
链接
代码
没加输入挂,334ms 加了输入挂64ms 。。 再一次感叹输入挂的强大。其实仔细想一下这道题的输入量 ,m*t*2==2*6000*t ;也就至少到了1e5级了。一般到1e5级的输入量,用输入挂,效果就很显著。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<iostream>
#include<vector>
using namespace std;
const int MAXN= 1001+10;
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch>'9'||ch<'0'){ if(ch=='-') f=-1; ch=getchar();}
while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}
return x*f;
} // 输入挂
/*------------------------------------*/
struct Edge {
int from,to,next;
}edge[6000+10];
int head[MAXN],top;
int n,m;
void init(){
memset(head,-1,sizeof(head));
top=0;
}
void addedge(int a,int b){
Edge e={a,b,head[a]};
edge[top]=e;head[a]=top++;
}
void getmap(){
int a,b;
while(m--){
a=read();b=read();
addedge(a,b);
}
}
int low[MAXN],dfn[MAXN];
int scc_cnt,sccno[MAXN];
stack<int>S;int Instack[MAXN];
vector<int>G[MAXN];
int dfs_clock;
void tarjan(int now,int par){
low[now]=dfn[now]=++dfs_clock;
S.push(now);Instack[now]=1;
for(int i=head[now];i!=-1;i=edge[i].next){
Edge e=edge[i];
if(!dfn[e.to]){
tarjan(e.to,now);
low[now]=min(low[now],low[e.to]);
}else if(Instack[e.to])
low[now]=min(low[now],dfn[e.to]);
}
if(low[now]==dfn[now]) {
scc_cnt++;
for(;;){
int nexts=S.top();S.pop();Instack[nexts]=0;
sccno[nexts]=scc_cnt;
if(nexts==now) break;
}
}
}
void find_cut(int le,int ri){
memset(low,0,sizeof(low));
memset(dfn,0,sizeof(dfn));
memset(Instack,0,sizeof(Instack));
memset(sccno,0,sizeof(sccno));
dfs_clock=scc_cnt=0;
for(int i=le;i<=ri;i++){
if(!dfn[i]) tarjan(i,-1);
}
}
int in[MAXN];
void suodian(){
for(int i=1;i<=scc_cnt;i++) {
in[i]=0;G[i].clear();
}
for(int i=0;i<top;i++){
Edge e=edge[i];
int now=sccno[e.from];
int nexts=sccno[e.to];
if(now!=nexts){
G[now].push_back(nexts);
in[nexts]++;
}
}
}
queue<int>Q;
bool topo(){
while(!Q.empty()) Q.pop();
int num=0;
for(int i=1;i<=scc_cnt;i++){
if(!in[i]) {
Q.push(i);
num++;
if(num>1) return false;
}
}
int k=0;
while(!Q.empty()){
int now=Q.front();Q.pop();num=0;k++;
for(int i=0;i<G[now].size();i++){
int v=G[now][i];
if(--in[v]==0){
Q.push(v);
num++;
if(num>1) return false;
}
}
}
return k==scc_cnt;
}
void solve(){
find_cut(1,n);
suodian();
if(scc_cnt==1) puts("Yes");
else
puts(topo()?"Yes":"No");
}
int main(){
int t;t=read();while(t--){
n=read();m=read();
init();
getmap();
solve();
}
return 0;
}