hdu 1932(spfa)

XYZZY

Description

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
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
题意:某人从 1走到 n,初始值为100，经过的每个点都有一个能量值,他走到每个点时会增加或者减少这个点的能量值,但是都不能小于 0 ，问他最后能否走到第n个点?题解:spfa 判断正环,如果某个点的入队次数>n，那么则证明这个数赋值 INF ，然后这个点就不能进入了，然后每一次要判断是否大于 0 

#include <iostream>
#include <cstdio>
#include <string.h>
#include <queue>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long LL;
const int INF = 999999999;
const int N = 205;
const int M = N*N;
struct Edge{
    int v,next;
}edge[M];
int head[N];
int tot;
int n;
int val[N];
void addEdge(int u,int v,int &k){
    edge[k].v = v,edge[k].next = head[u],head[u] = k++;
}
void init(){
    memset(head,-1,sizeof(head));
    tot = 0;
}
int low[N],time[N],pre[N];
bool vis[N];
bool spfa(int s){
    for(int i=1;i<=n;i++){
        vis[i] = false;
        low[i] = -INF;
        time[i] = 0;
        pre[i] = i;
    }
    low[s] = 100;
    time[s]++;
    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        if(time[u]>n) low[u] = INF;
        for(int k = head[u];k!=-1;k=edge[k].next){
            int v = edge[k].v;
            if(low[v]<low[u]+val[v]&&low[u]+val[v]>0){
                low[v] = low[u]+val[v];
                if(!vis[v]&&time[v]<=n){
                    vis[v] = true;
                    q.push(v);
                    time[v]++;
                }
            }
        }
    }
    int temp = n;
    if(low[n]<=0) return false;
    return true;
}
int main(){
    while(scanf("%d",&n)!=EOF,n!=-1){
        init();
        int num;
        for(int u=1;u<=n;u++){
            scanf("%d%d",&val[u],&num);
            for(int j=1;j<=num;j++){
                int v;
                scanf("%d",&v);
                addEdge(u,v,tot);
            }
        }
        if(spfa(1)) printf("winnable\n");
        else printf("hopeless\n");
    }
    return 0;
}