XYZZY UVA10557

Problem D: XYZZY

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ADVENT: /ad�vent/, n.

The prototypical computer adventure game, first designed by Will Crowther on the PDP-10 in the mid-1970s as an attempt at computer-refereed fantasy gaming, and expanded into a puzzle-oriented game by Don Woods at Stanford in 1976. (Woods had been one of the authors of INTERCAL.) Now better known as Adventure or Colossal Cave Adventure, but the TOPS-10 operating system permitted only six-letter filenames in uppercase. See also vadding, Zork, and Infocom.

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It has recently been discovered how to run open-source software on the Y-Crate gaming device. A number of enterprising designers have developedAdvent-style games for deployment on the Y-Crate. Your job is to test a number of these designs to see which are winnable.

Each game consists of a set of up to 100 rooms. One of the rooms is thestartand one of the rooms is thefinish. Each room has anenergy valuebetween -100 and +100. One-way doorways interconnect pairs of rooms.

The player begins in the start room with 100energy points. She may pass through any doorway that connects the room she is in to another room, thus entering the other room. The energy value of this room is added to the player's energy. This process continues until she wins by entering the finish room or dies by running out of energy (or quits in frustration). During her adventure the player may enter the same room several times, receiving its energy each time.

The input consists of several test cases. Each test case begins withn, the number of rooms. The rooms are numbered from 1 (the start room) ton(the finish room). Input for thenrooms follows. The input for each room consists of one or more lines containing:

• the energy value for roomi
• the number of doorways leaving roomi
• a list of the rooms that are reachable by the doorways leaving roomi

The start and finish rooms will always have enery level 0.A line containing -1 follows the last test case.

In one line for each case, output "winnable" if it is possible for the player to win, otherwise output "hopeless".

Sample Input

5
0 1 2
-60 1 3
-60 1 4
20 1 5
0 0
5
0 1 2
20 1 3
-60 1 4
-60 1 5
0 0
5
0 1 2
21 1 3
-60 1 4
-60 1 5
0 0
5
0 1 2
20 2 1 3
-60 1 4
-60 1 5
0 0
-1

Output for Sample Input

hopeless
hopeless
winnable
winnable

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G. V. Cormack

这道题用到了一个独特的算法SPFA，同时需要判断是否存在正环，刚开始的无从下手，后来参考别人的代码，就看明白了。

#include<iostream>
#include<cstring>
#include<queue>

using namespace std;

#define maxn 110

int grid[maxn][maxn],val[maxn][maxn],num[maxn][maxn];
int vis[maxn],dis[maxn];
int n,tag,s;
queue<int> t;

void dfs(int pos)
{
    if(pos==n)
    {
        tag=1;
        return;
    }
    vis[pos]=1;
    for(int i=1;i<=n;i++)
    {
        if(grid[pos][i]&&vis[i]==0&&tag==0)
        {
            dfs(i);
        }
    }
}


void spfa()
{
    while(!t.empty())
        t.pop();
    memset(vis,0,sizeof(vis));
    //memset(dis,0,sizeof(dis));
    vis[1]=1,dis[1]=100;
    s=0;
    t.push(1);
    while(!t.empty())
    {
        int x=t.front();
        t.pop();
        vis[x]=0;
        for(int i=1;i<=n;i++)
        {
            if(grid[x][i])
            {
                if(dis[i]<dis[x]+val[x][i])
                {
                    num[x][i]++;
                    dis[i]=dis[x]+val[x][i];
                    if(num[x][i]>=n)//判断回路
                    {
                        s=i;
                        return;
                    }
                    if(vis[i]==0&&dis[i]>0)
                    {
                        t.push(i);
                        vis[i]=1;
                    }
                }
            }
        }
    }
}

int main()
{
    while(cin>>n&&n!=-1)
    {
        memset(val,0,sizeof(val));
        memset(grid,0,sizeof(grid));
        memset(num,0,sizeof(num));
        //memset(vis,0,sizeof(vis));
        memset(dis,0,sizeof(dis));
        int i,j,k,v,l,r;
        for(i=1;i<=n;i++)
        {
            cin>>v>>k;
            for(j=1;j<=k;j++)
            {
                cin>>r;
                grid[i][r]=1;
                val[i][r]=v;
            }
        }
        spfa();
        if(!s)
        {
            tag=1;
            if(dis[n]<=0)
                tag=0;
        }
        else
        {
            tag=0;
            memset(vis,0,sizeof(vis));
            dfs(s);//判断从回路中某点出发能否到达终点
        }
        if(tag) cout<<"winnable"<<endl;
        else cout<<"hopeless"<<endl;
    }
    return 0;
}