The Tag Game CodeForces

首先 想要尽可能拖延游戏时间 Bob必须躲在一个离Alice尽可能远的叶节点(度为1)

而怎样知道两人分别到每个叶节点的距离呢?

分别以两人位置为根 递归遍历整张树图 节点在第几层递归遇到 它就在第几层 离根节点的距离就是多少

再以与Alice距离为依据 将所有叶节点排序 距离长者优先考虑

但可能在Bob前往该地时会被Alice截击 也就是说Alice离该点更近

这条路行不通 考虑离Alice第二远的叶节点 总能找到答案

#include <stdio.h>#include <queue>#include <vector>using namespace std;struct node{    friend bool operator< (node n1,node n2)    {        return n1.val<n2.val;    }    int id;    int val;};priority_queue <struct node> que;vector <int> edge[400001];int degree[200001],tem[200001],book[200001];int dis1[200001],last1[200001];int dis2[200001],last2[200001];int n,m,s,num;void safari1(int cur,int step);void safari2(int cur,int step);int main(){    struct node t;    int i,a,b,ans;    while(scanf("%d%d",&n,&s)!=EOF)    {        m=n-1;        for(i=1;i<=n;i++)        {            edge[i].clear();            degree[i]=0;        }        while(!que.empty())        {            que.pop();        }        for(i=1;i<=m;i++)        {            scanf("%d%d",&a,&b);            edge[a].push_back(b);            edge[b].push_back(a);            degree[a]++,degree[b]++;        }        num=0;        for(i=1;i<=n;i++)        {            book[i]=0;        }        dis1[1]=0,last1[1]=-1,book[1]=1;        safari1(1,1);        for(i=1;i<=n;i++)        {            book[i]=0;        }        dis2[s]=0,last2[s]=-1,book[s]=1;        safari2(s,1);        for(i=1;i<=num;i++)        {            if(dis1[tem[i]]>dis1[s])            {                t.id=tem[i];                t.val=dis1[tem[i]];                que.push(t);            }        }        ans=dis1[s];        while(!que.empty())        {            t=que.top();            que.pop();            if(dis2[t.id]<dis1[t.id])            {                ans=dis1[t.id];                break;            }        }        printf("%d\n",ans*2);    }    return 0;}void safari1(int cur,int step){    struct node t;    int i;    if(degree[cur]==1)////    {        num++;        tem[num]=cur;        if(cur!=1) return;    }    for(i=0;i<edge[cur].size();i++)    {        if(book[edge[cur][i]]==0)        {            dis1[edge[cur][i]]=step,last1[edge[cur][i]]=cur,book[edge[cur][i]]=1;            safari1(edge[cur][i],step+1);        }    }    return;}void safari2(int cur,int step){    int i;    if(cur!=s&°ree[cur]==1)////    {        return;    }    for(i=0;i<edge[cur].size();i++)    {        if(book[edge[cur][i]]==0)        {            dis2[edge[cur][i]]=step,last2[edge[cur][i]]=cur,book[edge[cur][i]]=1;            safari2(edge[cur][i],step+1);        }    }    return;}