2017 Multi-University Training Contest

FFF at Valentine

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 304    Accepted Submission(s): 143

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!

 

• 题意：

  • 给定一个有向图，n个点，m条边，无自环，无重边
  • 问：是否任意两点A,B，满足AB连通

• 规模：

  • 2<=n<=1000,m<=6000
  • T<120

• 类型：

  • 有向强连通，拓扑排序

• 分析：

  • 有向图缩环后，拓扑排序一下，如果入度为0的点有多个，就无法连通

#include <iostream>#include <bits/stdc++.h>using namespace std;const int MAXN = 1010;const int MAXM = 6010;struct Edge{    int to, next;} edge[MAXM], edge2[MAXM];int head[MAXN], head2[MAXN], tot, tot2;int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN];int Index, top;int scc;bool Instack[MAXN];int num[MAXN];int in[MAXN], out[MAXN];void addedge(int u, int v){    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}void addedge2(int u, int v){    edge2[tot2].to = v;    edge2[tot2].next = head2[u];    head2[u] = tot2++;}void Tarjan(int u){    int v;    Low[u] = DFN[u] = ++Index;    Stack[top++] = u;    Instack[u] = true;    for (int i = head[u]; i != -1; i = edge[i].next)    {        v = edge[i].to;        if (!DFN[v])        {            Tarjan(v);            if (Low[u] > Low[v])            {                Low[u] = Low[v];            }        }        else if (Instack[v] && Low[u] > DFN[v])        {            Low[u] = DFN[v];        }    }    if (Low[u] == DFN[u])    {        scc++;        do        {            v = Stack[--top];            Instack[v] = false;            Belong[v] = scc;            num[scc]++;        }        while (v != u);    }}void solve(int N){    memset(DFN, 0, sizeof(DFN));    memset(Instack, false, sizeof(Instack));    memset(num, 0, sizeof(num));    Index = scc = top = 0;    for (int i = 1; i <= N; i++)    {        if (!DFN[i])        {            Tarjan(i);        }    }}bool map2[MAXN][MAXN];void build(int n){    memset(map2, false, sizeof(map2));    memset(in, 0, sizeof(in));    memset(out, 0, sizeof(out));    memset(head2, -1, sizeof(head2));    tot2 = 0;    for (int i = 1; i <= n; i++)    {        for (int j = head[i]; j != -1; j = edge[j].next)        {            int v = edge[j].to;            int a = Belong[i];            int b = Belong[v];            if (a == b)            {                continue;            }            if (!map2[a][b])            {                addedge2(a, b);                map2[a][b] = true;                in[b]++;                out[a]++;            }        }    }}void init(){    tot = 0;    memset(head, -1, sizeof(head));}bool Top(){    queue<int >q;    while (!q.empty())    {        q.pop();    }    for (int i = 1; i <= scc; i++)    {        if (in[i] == 0)        {            q.push(i);        }    }    while (!q.empty())    {        if (q.size() != 1)        {            return false;        }        int u = q.front();        q.pop();        for (int i = 1; i <= scc; i++)        {            if (map2[u][i] == true)            {                in[i]--;                if (in[i] == 0)                {                    q.push(i);                }            }        }    }    return true;}int n, m;int main(){    int T;    scanf("%d", &T);    while (T--)    {        scanf("%d%d", &n, &m);        init();        for (int i = 0; i < m; i++)        {            int u, v;            scanf("%d%d", &u, &v);            addedge(u, v);        }        solve(n);        build(n);        if (!Top())        {            printf("Light my fire!\n");        }        else        {            printf("I love you my love and our love save us!\n");        }    }    return 0;}