网络寻路（dfs或者枚举）

这个题一开始用了dfs，因为蓝桥杯的数据太水了，然后就过了。但是之前做过类似的题，用枚举来做的，然后就是可以避免超时
//dfs

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <vector>#include <set>#include <map> #include <sstream>using namespace std;/* run this program using the console pauser or add your own getch, system("pause") or input loop */struct node{    int to,pre;};int h=0,ans=0;node e[200010];int head[10005];void add(int a,int b){    e[h].to=b;e[h].pre=head[a];head[a]=h;    h++;}int vis[10005];int start=0;void dfs(int bg,int time1){    if(time1==3){        ans++;        return ;    }    for(int i=head[bg];i>-1;i=e[i].pre)    {        node t=e[i];        if(time1==2&&t.to==start)         {          ans++;          continue;}        if(!vis[t.to])        {            vis[t.to]=1;            dfs(t.to,time1+1);            vis[t.to]=0;        }    }}int main(int argc, char *argv[]) {    int n,m;    scanf("%d %d",&n,&m);    memset(head,-1,sizeof(head));    for(int i=0;i<m;i++)    {        int a,b;        scanf("%d %d",&a,&b);        add(a,b);        add(b,a);    }    for(int i=1;i<=n;i++)    {        start=i;        memset(vis,0,sizeof(vis));        vis[i]=1;        dfs(i,0);    }    printf("%d\n",ans);    return 0;}

//枚举

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <vector>#include <set>#include <map> #include <sstream>using namespace std;/* run this program using the console pauser or add your own getch, system("pause") or input loop */struct node{    int to,pre;};set<pair<int ,int > > t;int h=0,ans=0;node e[200010];int head[10005];void add(int a,int b){    e[h].to=b;e[h].pre=head[a];head[a]=h;    h++;}int start=0;int main(int argc, char *argv[]) {    int n,m;    scanf("%d %d",&n,&m);    memset(head,-1,sizeof(head));    for(int i=0;i<m;i++)    {        int a,b;        scanf("%d %d",&a,&b);        add(a,b);        add(b,a);        t.insert(pair<int ,int > (a,b));    }    int sum=0;    set<pair<int ,int > > ::iterator it;    for(it=t.begin();it!=t.end();it++)    {           pair< int,int > temp=*it;        int a=temp.first,b=temp.second;//这个.first不要记错啦啦啦        int ans1=0,ans2=0;        for(int i=head[a];i>-1;i=e[i].pre)             if(e[i].to!=b) ans1++;        for(int i=head[b];i>-1;i=e[i].pre)             if(e[i].to!=a) ans2++;        sum+=ans1*ans2;    }    printf("%d\n",sum*2);    return 0;}