HDU 1317 XYZZY

XYZZY

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3608    Accepted Submission(s): 1018

Problem Description

It has recently been discovered how to run open-source software on the Y-Crate gaming device. A number of enterprising designers have developed Advent-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 the start and one of the rooms is the finish. Each room has an energy value between -100 and +100. One-way doorways interconnect pairs of rooms. 

The player begins in the start room with 100 energy 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. 

 

Input

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

the energy value for room i 
the number of doorways leaving room i 
a list of the rooms that are reachable by the doorways leaving room i 
The start and finish rooms will always have enery level 0. A line containing -1 follows the last test case. 

 

Output

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

 

Sample Output

hopeless hopeless winnable winnable

 

Source

University of Waterloo Local Contest 2003.09.27

 

Recommend

Eddy

 

这个题目意思就是

给你一堆房间 每个房间有一个值，进入房间时，你的精力+该房间的值，给你房间之间的单向连通关系，你初始有100精力，问你能不能走到最后一个房间。

一开始感觉没啥难的，就是跑spfa，如果有正环，原点到该点的距离设为正无穷

但是一直wa

后来比较代码发现 想法是没错，但是spfa有环的时候写错了！

原来是这么写的

                if(!flag[v])
                {
                    if(tims[v]<=m)
                    {
                        tims[v]++;
                        q.push(v);
                        flag[v]=1;
                    }
                    else
                    {
                        d[v]=inf;
                        flag[v]=1;
                    }
                }

这样存在的问题就是

你判断出来了正环，值也赋值了，但是这个点没有被加入到队列里用来更新后面的点！

简直傻狗！！！

所以改一下代码就好了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
int w[111];
int d[111];
int m,x,y;
int n;
struct self
{
    int x,y,w,nxt;
}s[111111];
int fst[111];
int ans;
queue<int>q;
int tims[111];
int flag[111];
void add(int x,int y)
{
    n++;
    s[n].x=x;
    s[n].y=y;
    s[n].nxt=fst[x];
    fst[x]=n;
}


void spfa()
{
    while(!q.empty())q.pop();
    memset(flag,0,sizeof(flag));
    memset(tims,0,sizeof(tims));
    for(int i=1;i<=m;i++)
        d[i]=-inf;
    q.push(1);
    d[1]=100;
    flag[1]=1;
    tims[1]=1;
    int ok=0;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        flag[u]=0;
        if(d[u]<=0)continue;
        //cout<<"d["<<u<<"]="<<d[u]<<endl;
        for(int e=fst[u];e!=-1;e=s[e].nxt)
        {
            int v=s[e].y;
            //cout<<" d["<<v<<"]="<<d[v]<<" w="<<w[v]<<endl;
            if(d[v]<d[u]+w[v])
            {
                d[v]=d[u]+w[v];
                if(!flag[v])
                {
                    if(tims[v]<m)
                    {
                        tims[v]++;
                        q.push(v);
                        flag[v]=1;
                    }
                    else
                        if(tims[v]==m)
                        {
                            d[v]=inf;
                            flag[v]=1;
                            q.push(v);
                            tims[v]++;
                        }
                }
            }
        }
    }
    if(d[m]>0)
        printf("winnable
");
    else
        printf("hopeless
");
}
int main()
{
    while(scanf("%d",&m)==1 && m!=-1)
    {
        memset(fst,-1,sizeof(fst));
        n=0;
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&w[i]);
            scanf("%d",&x);
            for(int j=1;j<=x;j++)
            {
                scanf("%d",&y);
                add(i,y);
            }
        }
        spfa();
    }
    return 0;
}