题意:一个有向图,无自环,无重边,让你判断这个图内的任意两点是否有路;
解题思路:首先,判断两个点是否可达一般用出入度来判断,如果在拓扑排序中同时有两个及以上入度同时为零的点,那么,这些入度的为零的点肯定不可达,因为没有路径指向它;然后就是简化图了,一个环的点肯定可达,所以缩下点,再拓扑排序下;
代码:
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<cstdio>
#include<cstring>
#define maxn 10005
using namespace std;
struct Edge
{int to;int next;
}edge[maxn];
struct node
{int x;int y;
}a[maxn];
int low[maxn];
int dfn[maxn];
int instack[maxn];
int visit[maxn];
int sccno[maxn];
int cnt;
int step;
int index;
int scc_cnt;
int head[maxn];
int indeg[maxn];
int indegree[maxn];
vector<int>scc[maxn];
void add(int u,int v)
{edge[cnt].next=head[u];edge[cnt].to=v;head[u]=cnt++;
}
void tarjan(int u)
{low[u]=dfn[u]=++step;instack[++index]=u;visit[u]=1;for(int i=head[u];i!=-1;i=edge[i].next){if(!dfn[edge[i].to]){tarjan(edge[i].to);low[u]=min(low[u],low[edge[i].to]);}else if(visit[edge[i].to]){low[u]=min(low[u],dfn[edge[i].to]);}}if(low[u]==dfn[u]){scc_cnt++;scc[scc_cnt].clear();do{scc[scc_cnt].push_back(instack[index]);sccno[instack[index]]=scc_cnt;visit[instack[index]]=0;index--;}while(u!=instack[index+1]);}return;
}
void init()
{memset(head,-1,sizeof(head));cnt=step=index=0;scc_cnt=0;memset(visit,0,sizeof(visit));memset(low,0,sizeof(low));memset(dfn,0,sizeof(dfn));memset(indegree,0,sizeof(indegree));
}
int topusort()
{int flag=0;queue<int>q;while(!q.empty())q.pop();for(int i=1;i<=scc_cnt;i++){indeg[i]=indegree[i];if(indeg[i]==0)q.push(i);}while(!q.empty()){if(q.size()>=2)flag=1;int temp=q.front();q.pop();for(int j=0;j<scc[temp].size();j++){for(int i=head[scc[temp][j]];i!=-1;i=edge[i].next){if(sccno[edge[i].to]!=temp){indeg[sccno[edge[i].to]]--;if(indeg[sccno[edge[i].to]]==0)q.push(sccno[edge[i].to]);}}}}return flag;
}
int main()
{int n,m;int x,y;int t;scanf("%d",&t);while(t--){init();scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){scanf("%d%d",&a[i].x,&a[i].y);add(a[i].x,a[i].y);}for(int i=1;i<=n;i++)if(!dfn[i])tarjan(i);for(int i=1;i<=m;i++)if(sccno[a[i].x]!=sccno[a[i].y])indegree[sccno[a[i].y]]++;int ans=topusort();if(ans==1)printf("Light my fire!\n");elseprintf("I love you my love and our love save us!\n");}
}