HDU 6165 FFF at Valentine（tarjan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): 967    Accepted Submission(s): 466

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：

给你一个图，含环，让你判断图内的任意两点能从一个点到另一个点，单向即满足要求。

如果有不满足要求的，则不行。反之则行。

POINT：

先tarjan算法缩点，变成有向无环图。

开始拓扑排序，若途中有两个入度为0的点，则不满足要求。

排序可以结束则满足要求。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <stack>
using namespace std;
const int maxn=1100;
const int maxm=6600;
int low[maxn],DFN[maxn],instack[maxn],tm,cor[maxn],cornum;
vector<int> G[maxn],GG[maxn];
stack<int>q;
void tarjan(int u)
{
    DFN[u]=low[u]=++tm;
    instack[u]=1;
    q.push(u);
    for(int i=0;i<G[u].size();i++)
    {
        int v=G[u][i];
        if(DFN[v]==0)
        {
            tarjan(v);
            if(low[u]>low[v])   low[u]=low[v];
        }
        else if(instack[v]&&low[u]>DFN[v])
            low[u]=DFN[v];
    }
    if(DFN[u]==low[u])
    {
        cornum++;
        int v;
        do
        {
            v=q.top();
            cor[v]=cornum;
            instack[v]=0;
            q.pop();
        }while(v!=u);
    }
}
void init()
{
    for(int i=1;i<maxn;i++)
    {
        G[i].clear();
        GG[i].clear();
    }
    memset(low,0,sizeof low);
    memset(DFN,0,sizeof DFN);
    memset(instack,0,sizeof instack);
    memset(cor,0,sizeof cor);
    tm=0;
    cornum=0;
}
void add(int u,int v)
{
    G[u].push_back(v);
}
int ru[maxn];
bool Topo()
{
    int vis[maxn];
    memset(vis,0,sizeof vis);
    for(int i=1;i<=cornum;i++)
    {
        int k=0;
        for(int j=1;j<=cornum;j++)
        {
            if(!vis[j]&&ru[j]==0)
            {
                if(k!=0) return 0;
                k=j;
            }
        }
        vis[k]=1;
        for(int j=0;j<GG[k].size();j++)
        {
            ru[GG[k][j]]--;
        }
    }
    return 1;
}
int main()
{
    int T;
    scanf("%d",&T);
    int uu[maxm],vv[maxm];
    while(T--)
    {
        init();
        int n,m;scanf("%d %d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&uu[i],&vv[i]);
            add(uu[i],vv[i]);
        }
        for(int i=1;i<=n;i++)
            if(DFN[i]==0) tarjan(i);
      //  for(int i=1;i<=n;i++) printf("%d ",cor[i]);
        memset(ru,0,sizeof ru);
        for(int i=1;i<=m;i++)
        {
            if(cor[uu[i]]==cor[vv[i]]) continue;
            GG[cor[uu[i]]].push_back(cor[vv[i]]);
            ru[cor[vv[i]]]++;
        }
        if(Topo())
        {
            printf("I love you my love and our love save us!\n");
        }
        else printf("Light my fire!\n");
            
    }
}