题目传送门

  建图后，本题考察1到n这两个点是否联通，以及求最长路判断正环，floyd()判断连通性，spfa判断是否有正环即可。
  Talk is cheap.Show coder.

#include<cstdio>
#include<cstring>
#include<cctype>
#include<queue>
#include<algorithm>
#include<iostream>
using namespace std;
int dis[110],in[110],val[110];
bool maze1[110][110],maze2[110][110];
int n;
void Floyd()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
           for(int k=1;k<=n;k++)
               maze2[i][j]=(maze2[i][j]||(maze2[i][k]&&maze2[k][j]));
}

bool spfa()
{
    queue<int>Q;
    Q.push(1);
    memset(in,0,sizeof(in));
    memset(dis,0,sizeof(dis));
    dis[1]=100;
    while(!Q.empty())
    {
        int cur=Q.front();
        Q.pop();
        in[cur]++;
        if(in[cur]>=n)
            return maze2[cur][n];
        for(int next=1;next<=n;next++)
        {
            if(maze1[cur][next]&&dis[cur]+val[next]>dis[next]&&dis[cur]+val[next]>0)
            {
                Q.push(next);
                dis[next]=dis[cur]+val[next];
            }
        }
    }
    if(dis[n]>0)
        return true;
    return false;
}
int main()
{
    while(~scanf("%d",&n))
    {
        if(n==-1)break;
        memset(maze1,false,sizeof(maze1));
        memset(maze2,false,sizeof(maze2));
        for(int i=1;i<=n;i++)
        {
            int num,tmp;
            scanf("%d",&val[i]);
            scanf("%d",&num);
            for(int j=1;j<=num;j++)
            {
                scanf("%d",&tmp);
                maze1[i][tmp]=maze2[i][tmp]=true;
            }
        }
        Floyd();
        if(!maze2[1][n])
        {
            puts("hopeless");
        }
        else
        {
            if(spfa())
                puts("winnable");
            else
                puts("hopeless");
        }
    }
    return 0;
}