UVA 10557 XYZZY

题目如下：

Problem D: XYZZY

---

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.

---

It has recently been discovered how to run open-source software on theY-Crate gaming device. A number of enterprising designers have developedAdvent-style games for deployment on the Y-Crate. Your jobis 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 therooms 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 maypass through any doorway that connects the room she is in to another room, thusentering the other room. The energy value of this room is added tothe player's energy. This process continues until she wins by enteringthe finish room or dies by running out of energy (or quits in frustration). During her adventurethe player may enter the same room several times, receiving its energyeach 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) 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.

In one line for each case, output "winnable" if it is possible forthe 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

---

问一个点到另一个点是否可达，每个点有权值（能量），要求能量和始终大于0，并且可能有圈。在无圈的情况下用SPFA算法可以求得最长路径，只要最长路径大于0就可以判断为成功。在有圈情况下，SPFA可以判断出圈（节点总数大于N即说明有圈），如果产生的是正圈（负圈直接无视，因为不会傻到走负圈让能量减少），并且产生正圈的点在不考虑能量的情况下可以到达终点（直接DFS即可），则可以判断为成功。否则失败。因为其余情况要么是点根本就不能到达终点，要么是能量不满足。

AC的代码如下；

#include 
using namespace std;
int road[120][120];
int energy[120];
int vis[120];
int n;
int sumene[120];
int inq[120];
int sumq[120];
int flag;
void dfs(int u)
{
    if(vis[u])
        return;
    vis[u]=1;
    for(int i=1; i<=n; i++)
    {
        if(road[i][u]==1)
            dfs(i);
    }
}
int q[120];
void spfa()
{
    int front=0,rear=0;
    sumene[1]=100;
    q[rear++]=1;
    inq[1]=1;
    sumq[1]++;
    while(front!=rear)
    {
        int u=q[front++];
        if(front>n)
            front=0;
        inq[u]=0;
        for(int i=1; i<=n; i++)
        {
            if(road[u][i]&&vis[i]&&sumene[u]+energy[i]>sumene[i])
            {
                sumene[i]=sumene[u]+energy[i];
                if(!inq[i])
                {
                    q[rear++]=i;
                    if(rear>n)
                        rear=0;
                    inq[i]=1;
                    if(sumq[i]++>n)
                    {
                        flag=1;
                        break;
                    }
                }
                if(sumene[n]>0||flag)
                    break;
            }
        }
        if(sumene[n]>0||flag)
            break;
    }
    if(sumene[n]>0||flag)
        cout<<"winnable"<>n)
    {
        if(n==-1)
            break;
        memset(vis,0,sizeof vis);
        memset(road,0,sizeof road);
        memset(energy,0,sizeof energy);
        memset(sumene,0,sizeof sumene);
        memset(sumq,0,sizeof sumq);
        memset(inq,0,sizeof inq);
        memset(q,0,sizeof q);
        flag=0;
        for(int i=1; i<=n; i++)
        {
            int w;
            cin>>energy[i]>>w;
            for(int j=1; j<=w; j++)
            {
                int d;
                cin>>d;
                road[i][d]=1;
            }
        }
        dfs(n);
        if(!vis[1])
        {
            cout<<"hopeless"<