hdu6165 FFF at Valentine 2017多校第九场1005 dfs

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http://acm.split.hdu.edu.cn/showproblem.php?pid=6165

题意：给出一个有向图，要求图上任意两点是可达的(不需要相互可达)

题解：dfs，每次搜索每个结点的儿子然后标记连通即可。

代码：

#include<bits/stdc++.h>#define debug cout<<"aaa"<<endl#define d(a) cout<<a<<endl#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define lson l,mid,root<<1#define rson mid+1,r,root<<1|1#define MIN_INT (-2147483647-1)#define MAX_INT 2147483647#define MAX_LL 9223372036854775807i64#define MIN_LL (-9223372036854775807i64-1)using namespace std;const int N = 1000 + 5;const int mod = 1000000000 + 7;const double eps = 1e-8;int dp[N][N];//可达标记 bool vis[N];vector<int> son[N];void dfs(int u,int fa){dp[fa][u]=vis[u]=1;//记为可达 for(int i=0;i<son[u].size();i++){int v=son[u][i];if(!vis[v]){dfs(v,fa);}}}int main(){int t,m,n,u,v;bool flag;scanf("%d",&t);while(t--){flag=1,mem(dp,0),mem(vis,0);scanf("%d%d",&n,&m);for(int i=0;i<=n;i++){son[i].clear();}for(int i=1;i<=m;i++){scanf("%d%d",&u,&v);dp[u][v]=1;son[u].push_back(v);}for(int i=1;i<=n;i++){vis[i]=1;dfs(i,i);mem(vis,0);}for(int i=1;i<=n;i++){for(int j=1;j<i;j++){if(dp[i][j]||dp[j][i]) continue;flag=0;}}if(flag){puts("I love you my love and our love save us!");}else{puts("Light my fire!");}}return 0;}

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