FFF at Valentine(强连通分量缩点+拓扑排序)

FFF at Valentine

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 730    Accepted Submission(s): 359


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

 

//题意:给出一个有向图，问是否任意两点都可以有，至少从其中一点到另一点可行的路径

//题解:首先想到的是好像是问是否是强连通图，然后看清题后发现并不是，求出连通分量缩点后变为有向无环图后，只需要确定，有唯一的拓扑排序的结果即可

  1 # include <cstring>
  2 # include <cstdio>
  3 # include <cstdlib>
  4 # include <iostream>
  5 # include <vector>
  6 # include <queue>
  7 # include <stack>
  8 # include <map>
  9 # include <bitset>
 10 # include <sstream>
 11 # include <set>
 12 # include <cmath>
 13 # include <algorithm>
 14 # pragma  comment(linker,"/STACK:102400000,102400000")
 15 using namespace std;
 16 # define LL          long long
 17 # define pr          pair
 18 # define mkp         make_pair
 19 # define lowbit(x)   ((x)&(-x))
 20 # define PI          acos(-1.0)
 21 # define INF         0x3f3f3f3f3f3f3f3f
 22 # define eps         1e-8
 23 # define MOD         1000000007
 24 
 25 inline int scan() {
 26     int x=0,f=1; char ch=getchar();
 27     while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
 28     while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
 29     return x*f;
 30 }
 31 inline void Out(int a) {
 32     if(a<0) {putchar('-'); a=-a;}
 33     if(a>=10) Out(a/10);
 34     putchar(a%10+'0');
 35 }
 36 const int N = 1005;
 37 const int M = 6005;
 38 /**************************/
 39 struct Edge
 40 {
 41     int to;
 42     int nex;
 43 }edge[M*2];
 44 int n,m,realm,scc,Ddex;
 45 int hlist[N],hlist2[N];
 46 int dfn[N],low[N],belong[N];
 47 bool instk[N];
 48 stack<int> stk;
 49 int indu[N];
 50 
 51 void addedge(int u,int v)
 52 {
 53     edge[realm] = (Edge){v,hlist[u]};
 54     hlist[u]=realm++;
 55 }
 56 void addedge2(int u,int v)
 57 {
 58     edge[realm] = (Edge){v,hlist2[u]};
 59     hlist2[u]=realm++;
 60 }
 61 
 62 void Init_tarjan()
 63 {
 64     Ddex=0;scc=0;
 65     memset(dfn,0,sizeof(dfn));
 66     memset(instk,0,sizeof(instk));
 67 }
 68 
 69 void tarjan(int u)
 70 {
 71     dfn[u]=low[u]=++Ddex;
 72     stk.push(u); instk[u]=1;
 73     for (int i=hlist[u];i!=-1;i=edge[i].nex)
 74     {
 75         int v = edge[i].to;
 76         if (!dfn[v])
 77         {
 78             tarjan(v);
 79             low[u] =  min(low[u],low[v]);
 80         }
 81         else if (instk[v])
 82                 low[u] = min(low[u],dfn[v]);
 83     }
 84     if (dfn[u]==low[u])
 85     {
 86         scc++;
 87         while(1){
 88             int p = stk.top(); stk.pop();
 89             instk[p]=0;
 90             belong[p]=scc;
 91             if (u==p) break;
 92         }
 93     }
 94 }
 95 
 96 void build()
 97 {
 98     memset(hlist2,-1,sizeof(hlist2));
 99     memset(indu,0,sizeof(indu));
100     for (int i=1;i<=n;i++)
101     {
102         for (int j=hlist[i];j!=-1;j=edge[j].nex)
103         {
104             int x = belong[i];
105             int y = belong[edge[j].to];
106             if (x!=y)
107             {
108                 addedge2(x,y);
109                 indu[y]++;
110             }
111         }
112     }
113 }
114 
115 int topo()
116 {
117     queue<int> Q;
118     for (int i=1;i<=scc;i++)
119         if (indu[i]==0) Q.push(i);
120     if (Q.size()!=1) return 0;
121     while (!Q.empty())
122     {
123         int u = Q.front(); Q.pop();
124         for (int i=hlist2[u];i!=-1;i=edge[i].nex)
125         {
126             int v = edge[i].to;
127             indu[v]--;
128             if (indu[v]==0)
129                 Q.push(v);
130         }
131         if (Q.size()>1) return 0;
132     }
133     return 1;
134 }
135 
136 int main()
137 {
138     int T = scan();
139     while (T--)
140     {
141         memset(hlist,-1,sizeof(hlist));
142         realm=0;
143         scanf("%d%d",&n,&m);
144         for (int i=0;i<m;i++)
145         {
146             int u,v;
147             scanf("%d%d",&u,&v);
148             addedge(u,v);
149         }
150         Init_tarjan();
151         for (int i=1;i<=n;i++)
152             if (!dfn[i])
153                 tarjan(i);
154         build();//建新图
155         if (topo())//拓扑
156             printf("I love you my love and our love save us!\n");
157         else
158             printf("Light my fire!\n");
159     }
160     return 0;
161 }
View Code

 

posted @ 2017-08-23 17:30  happy_codes  阅读(274)  评论(0)  编辑  收藏  举报