hdu 6165 FFF at Valentine（传递闭包）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6165

FFF at Valentine

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 682    Accepted Submission(s): 334

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

 

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解析：floyd求传递闭包，统计没有关系的数量即可。bfs也可以 ,但这个题卡时间，floyd会T，所以就用bfs，直接板子

代码：

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1000+9;const int INF = 0x3f3f3f3f;vector<int> mp[N];int g[N][N];int n, m;bool judge(){    for(int i = 1; i <= n; i++)    {        for(int j = i + 1; j <= n; j++)            if(!(g[i][j] || g[j][i])) return false;    }    return true;}void bfs(int s){    queue<int> q;    q.push(s);    g[s][s] = 1;    while(!q.empty())    {        int u = q.front(); q.pop();        for(int i = 0; i < mp[u].size(); i++)        {            int v = mp[u][i];            if(!g[s][v]) q.push(v);            g[s][v] = 1;        }    }}void solve(){    memset(g, 0, sizeof(g));    for(int i = 1; i <= n; i++) bfs(i);}int main(){    int t, u, v;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &n, &m);        for(int i = 1; i <= n; i++) mp[i].clear();        while(m--)        {            scanf("%d%d", &u, &v);            mp[u].push_back(v);        }        solve();        if(judge()) puts("I love you my love and our love save us!");        else puts("Light my fire!");    }    return 0;}