[HDU 6165]FFF at Valentine

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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

35 51 22 32 43 54 53 31 22 33 15 51 22 33 13 44 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：求问一个有向图中任意两个点对（i,j），是否符合其中的一个点能够到达另一个点。

题解：tarjan缩环，然后在DAG上进行拓扑排序，如果每层拓扑进入队列中的点数大于1，则这些点不可能互相到达。

（还是犯傻了，数组没开够……RE无数次，之前还写的dfs，脑子估计是不存在的。）

#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<string>#include<stack>#include<queue>#define LiangJiaJun main#define ll long longusing namespace std;int T;queue<int>q;struct edge{    int to,nt;}e[12004],ed[12004];int n,m,h[2004],ne,scc,belong[2004],dfn[2004],low[2004];int cnt,l[2004],ru[2004];bool inq[2004];stack<int>st;void add(int u,int v){e[++ne].to=v;e[ne].nt=h[u];h[u]=ne;}void add2(int u,int v){ed[++cnt].to=v;ed[cnt].nt=l[u];l[u]=cnt;}void tarjan(int x){     dfn[x]=low[x]=++cnt;     inq[x]=1;st.push(x);     for(int i=h[x];i;i=e[i].nt){         if(!dfn[e[i].to]){             tarjan(e[i].to);             low[x]=min(low[x],low[e[i].to]);         }         else if(inq[e[i].to])low[x]=min(low[x],dfn[e[i].to]);     }     if(dfn[x]==low[x]){        ++scc;int now=0;        while(now!=x){            now=st.top();st.pop();            inq[now]=0;belong[now]=scc;        }     }}void rebuild(){     cnt=1;     for(int i=1;i<=n;i++){         for(int j=h[i];j;j=e[j].nt){             if(belong[e[j].to]!=belong[i]){                add2(belong[i],belong[e[j].to]);                ru[belong[e[j].to]]++;             }         }     }}int w33ha(){    memset(belong,0,sizeof(belong));    memset(h,0,sizeof(h));    memset(ru,0,sizeof(ru));    memset(inq,0,sizeof(inq));    memset(dfn,0,sizeof(dfn));    memset(low,0,sizeof(low));    memset(l,0,sizeof(l));    while(!q.empty())q.pop();    scc=0;ne=1;cnt=0;    scanf("%d%d",&n,&m);    for(int i=1;i<=m;i++){        int u,v;        scanf("%d%d",&u,&v);        add(u,v);    }    for(int i=1;i<=n;i++)if(!dfn[i])tarjan(i);    rebuild();    for(int i=1;i<=scc;i++)if(ru[i]==0)q.push(i);    while(!q.empty()){        if(q.size()>1)return puts("Light my fire!"),0;        int x=q.front();q.pop();        for(int i=l[x];i;i=ed[i].nt){            --ru[ed[i].to];            if(!ru[ed[i].to])q.push(ed[i].to);        }    }    puts("I love you my love and our love save us!");    return 0;}int LiangJiaJun (){    scanf("%d",&T);    while(T--)w33ha();    return 0;}