FFF at Valentine

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1060    Accepted Submission(s): 506

Problem Description

At
Valentine‘s eve, Shylock and Lucar were enjoying their time as any
other couples. Suddenly, LSH, Boss of FFF Group caught both of them, and
locked them into two separate cells of the jail randomly. But as the
saying goes: There is always a way out , the lovers made a bet with LSH:
if either of them can reach the cell of the other one, then LSH has to
let them go.
The jail is formed of several cells and each cell has
some special portals connect to a specific cell. One can be transported
to the connected cell by the portal, but be transported back is
impossible. There will not be a portal connecting a cell and itself, and
since the cost of a portal is pretty expensive, LSH would not tolerate
the fact that two portals connect exactly the same two cells.
As an
enthusiastic person of the FFF group, YOU are quit curious about whether
the lovers can survive or not. So you get a map of the jail and decide
to figure it out.

Input

?Input starts with an integer T (T≤120), denoting the number of test cases.
?For each case,
First line is two number n and m, the total number of cells and portals in the jail.(2≤n≤1000,m≤6000)
Then next m lines each contains two integer u and v, which indicates a portal from u to v.

Output

If the couple can survive, print “I love you my love and our love save us!”
Otherwise, print “Light my fire!”

Sample Input

3
5 5
1 2
2 3
2 4
3 5
4 5

3 3
1 2
2 3
3 1

5 5
1 2
2 3
3 1
3 4
4 5

Sample Output

Light my fire!
I love you my love and our love save us!
I love you my love and our love save us!

Source

2017 Multi-University Training Contest - Team 9

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6165

分析：缩点为DAG，则如果在拓扑序中出现了有两个及以上入度为0的点则不合法

下面给出AC代码：

  1 #include <iostream>
  2 #include <bits/stdc++.h>
  3 using namespace std;
  4 const int MAXN=1010;
  5 const int MAXM=6010;
  6 struct Edge{
  7     int to,next;
  8 }edge[MAXM],edge2[MAXM];
  9 int head[MAXN],head2[MAXN],tot,tot2;
 10 int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
 11 int Index,top;
 12 int scc;
 13 bool Instack[MAXN];
 14 int num[MAXN];
 15 int in[MAXN],out[MAXN];
 16 void addedge(int u,int v){
 17     edge[tot].to=v;edge[tot].next=head[u];head[u]=tot++;
 18 }
 19 void addedge2(int u,int v){
 20     edge2[tot2].to=v;edge2[tot2].next=head2[u];head2[u]=tot2++;
 21 }
 22 void Tarjan(int u){
 23     int v;
 24     Low[u]=DFN[u]=++Index;
 25     Stack[top++]=u;
 26     Instack[u]=true;
 27     for(int i=head[u];i!=-1;i=edge[i].next){
 28         v=edge[i].to;
 29         if(!DFN[v]){
 30             Tarjan(v);
 31             if(Low[u]>Low[v])Low[u]=Low[v];
 32         }
 33         else if(Instack[v]&&Low[u]>DFN[v])
 34             Low[u]=DFN[v];
 35     }
 36     if(Low[u]==DFN[u]){
 37         scc++;
 38         do{
 39             v=Stack[--top];
 40             Instack[v]=false;
 41             Belong[v]=scc;
 42             num[scc]++;
 43         }
 44         while(v!=u);
 45     }
 46 }
 47 void solve(int N){
 48     memset(DFN,0,sizeof(DFN));
 49     memset(Instack,false,sizeof(Instack));
 50     memset(num,0,sizeof(num));
 51     Index=scc=top=0;
 52     for(int i=1;i<=N;i++){
 53         if(!DFN[i])
 54             Tarjan(i);
 55     }
 56 }
 57
 58 bool map2[MAXN][MAXN];
 59 void build(int n){
 60     memset(map2,false,sizeof(map2));
 61     memset(in,0,sizeof(in));
 62     memset(out,0,sizeof(out));
 63     memset(head2,-1,sizeof(head2));tot2=0;
 64     for(int i=1;i<=n;i++){
 65         for(int j=head[i];j!=-1;j=edge[j].next){
 66             int v=edge[j].to;
 67             int a=Belong[i];
 68             int b=Belong[v];
 69             if(a==b)continue;
 70             if(!map2[a][b]){
 71                 addedge2(a,b);
 72                 map2[a][b]=true;
 73                 in[b]++;out[a]++;
 74             }
 75         }
 76     }
 77 }
 78
 79 void init(){
 80     tot=0;
 81     memset(head,-1,sizeof(head));
 82 }
 83
 84 bool Top(){
 85     queue<int >q;
 86     while(!q.empty())q.pop();
 87     for(int i=1;i<=scc;i++){
 88         if(in[i]==0)q.push(i);
 89     }
 90
 91     while(!q.empty()){
 92         if(q.size()!=1)return false;
 93         int u=q.front();
 94         q.pop();
 95         for(int i=1;i<=scc;i++){
 96             if(map2[u][i]==true) {
 97                 in[i]--;
 98                 if(in[i]==0)q.push(i);
 99             }
100         }
101     }
102     return true;
103 }
104
105 int n,m;
106 int main()
107 {
108     int T;
109     scanf("%d",&T);
110     while(T--){
111         scanf("%d%d",&n,&m);
112         init();
113         for(int i=0;i<m;i++){
114             int u,v;
115             scanf("%d%d",&u,&v);
116             addedge(u,v);
117         }
118         solve(n);
119         build(n);
120
121         if(!Top()){printf("Light my fire!\n");}
122         else printf("I love you my love and our love save us!\n");
123     }
124
125     return 0;
126 }