【链接】http://acm.hdu.edu.cn/showproblem.php?pid=6165
【题意】
一张有向图，n个点，m条边，保证没有重边和自环。询问任意两个点能否满足任何一方能够到达另外一方。
【题解】
用Tarjan算法,先把有向图的强连通分量缩成一个点,缩完点之后,剩下的就是一张有向无环图了.
对其进行拓扑排序.一定要唯一的拓扑排序才能够满足题目的要求.
也即,为一条链的时候.
一旦某个时刻做拓扑排序的队列大小大于1就输出无解
【错的次数】
0
【反思】
在这了写反思
【代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x)
#define oi(x) printf("%d",x)
#define ol(x) printf("%lld",x)
#define oc putchar(' ')
#define os(x) printf(x)
#define all(x) x.begin(),x.end()
#define Open() freopen("F:\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1e3;
vector <int> G[N+10],g[N+10];
int n,m,tot = 0,top = 0,dfn[N+10],low[N+10],z[N+10],totn,du[N+10],in[N+10];
int bh[N+10];
queue <int> dl;
void dfs(int x){
dfn[x] = low[x] = ++ tot;
z[++top] = x;
in[x] = 1;
int len = G[x].size();
rep1(i,0,len-1){
int y = G[x][i];
if (!dfn[y]){
dfs(y);
low[x] = min(low[x],low[y]);
}else
if (in[y] && dfn[y]<low[x]){
low[x] = dfn[y];
}
}
if (low[x]==dfn[x]){
int v = 0;
totn++;
while (v!=x){
v = z[top];
in[v] = 0;
bh[v] = totn;
top--;
}
}
}
bool ju(){
return ((int) dl.size()) > 1;
}
bool ok(){
while (!dl.empty()) dl.pop();
rep1(i,1,n)
if (du[i]==0)
dl.push(i);
while (!dl.empty()){
if (ju()) return false;
int x = dl.front();
dl.pop();
int len = g[x].size();
rep1(i,0,len-1){
int y = g[x][i];
du[y]--;
if (du[y]==0){
dl.push(y);
}
}
}
return true;
}
int main(){
//Open();
//Close();
int T;
ri(T);
while (T--){
ms(dfn,0);
ms(du,0);
ms(in,0);
tot = 0,totn = 0;
ri(n),ri(m);
rep1(i,1,n) G[i].clear(),g[i].clear();
rep1(i,1,m){
int x,y;
ri(x),ri(y);
G[x].pb(y);
}
rep1(i,1,n)
if (dfn[i]==0)
dfs(i);
rep1(i,1,n){
int len = G[i].size();
int xx = bh[i];
rep1(j,0,len-1){
int y = G[i][j];
int yy = bh[y];
if (xx!=yy){
g[xx].pb(yy);
du[yy]++;
}
}
}
n = totn;
if (!ok())
puts("Light my fire!");
else
puts("I love you my love and our love save us!");
}
return 0;
}