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;
}
