hdu 6165 FFF at Valentine（强连通分量缩点+拓扑排序）

FFF at Valentine

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!

题意：给你一个无自环、无重边的有向图，问图中任意两点之间是否都存在一条路径可达

思路：可能有环存在，因此我们可以先用tarjan求出强连通分量，然后缩点建立新图，再利用拓扑排序判断新图中的每一层是否只有一个入度为0的点（每一层只有一个入度为0的点时满足题意）

代码：

#include<bits/stdc++.h>using namespace std;const int maxn=1000+5;struct edge{    int v,next;} E[maxn*6];int DFN[maxn],LOW[maxn],Belong[maxn],Stack[maxn];int first[maxn],in[maxn];bool Instack[maxn];int n,m,len,top,Index,Bcnt;vector<int>p[maxn];void Init(){    memset(in,0,sizeof(in));    memset(DFN,0,sizeof(DFN));    memset(LOW,0,sizeof(LOW));    memset(Stack,0,sizeof(Stack));    memset(first,-1,sizeof(first));    memset(Belong,0,sizeof(Belong));    memset(Instack,false,sizeof(Instack));    len=top=Index=Bcnt=0;    for(int i=1; i<=n; ++i)        p[i].clear();}void add_edge(int u,int v){    E[len].v=v,E[len].next=first[u],first[u]=len++;}void tarjan(int u){    int v;    DFN[u]=LOW[u]=++Index;    Stack[++top]=u;    Instack[u]=true;    for(int i=first[u]; ~i; i=E[i].next)    {        v=E[i].v;        if(!DFN[v])        {            tarjan(v);            LOW[u]=min(LOW[u],LOW[v]);        }        else if(Instack[v])            LOW[u]=min(LOW[u],DFN[v]);    }    if(DFN[u]==LOW[u])    {        ++Bcnt;        do        {            v=Stack[top--];            Instack[v]=false;            Belong[v]=Bcnt;        }        while(u!=v);    }}bool judge(){    int num=0,u;    queue<int>q;    for(int i=1; i<=Bcnt; ++i)        if(!in[i])            ++num,q.push(i);    if(num>1)        return false;    while(!q.empty())    {        u=q.front();        q.pop(),num=0;        for(int i=0; i<p[u].size(); ++i)        {            --in[p[u][i]];            if(!in[p[u][i]])                ++num,q.push(p[u][i]);        }        if(num>1)            return false;    }    return true;}int main(){    int t,u,v;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        Init();        for(int i=0; i<m; ++i)        {            scanf("%d%d",&u,&v);            add_edge(u,v);        }        for(int i=1; i<=n; ++i)            if(!DFN[i])                tarjan(i);        for(int i=1; i<=n; ++i)        {            u=Belong[i];            for(int j=first[i]; ~j; j=E[j].next)            {                v=Belong[E[j].v];                if(u!=v)                    p[u].push_back(v),++in[v];            }        }        if(judge())            printf("I love you my love and our love save us!\n");        else            printf("Light my fire!\n");    }    return 0;}

tarjan算法求强连通分量：九野
tarjan算法模板：sentimental_dog