| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 3105 | Accepted: 887 |
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 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
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
//208K 157MS C++ 1444B 2013-11-23 17:33:33
/* 题意：
给出一个单向图，每个点有一个权值，每到一个点就要加上该点的权值，判断是否存在从
点1到点n的路径. 最短路径：
网上解法很多.这里是其中一种，spfa+floyd
题目解法思路是差不多的，就是先判断是否存在一条路径符合，如果不存在在判断途中是否有正环，
如果有正环就判断该环是否与点n连通，连通表示存在。
其中两部分都有很多算法可以解.有兴趣的同学可以自己试试.. 这里spfa判断是否存在最短路，floyd判断是否有是否连通
*/
#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
int n;
int g[][];
int w[];
int spfa()
{
queue<int>Q;
int max,now;
int vis[]={};
int in[]={};
int d[]={};
d[]=;
vis[]=;
Q.push();
in[]++;
while(!Q.empty()){ //spfa
int u=Q.front();
Q.pop();
vis[u]=;
if(in[u]>n) break;
for(int i=;i<=n;i++){
if(g[u][i] && d[i]<d[u]+w[i]){
d[i]=d[u]+w[i];
if(!vis[i]){
in[i]++;
Q.push(i);
vis[i]=;
}
}
}
}
if(d[n]>) return ; //存在最短路
else{
for(int k=;k<=n;k++) //floyd
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
if(g[i][k] && g[k][j])
g[i][j]=;
for(int i=;i<=n;i++)
if(in[i]>n && g[][i] && g[i][n]) //存在正环和连通
return ;
}
return ;
}
int main(void)
{
int m,to;
while(scanf("%d",&n),n!=-)
{
memset(g,,sizeof(g));
for(int i=;i<=n;i++){
scanf("%d%d",&w[i],&m);
while(m--){
scanf("%d",&to);
g[i][to]=;
}
}
if(spfa()) puts("winnable");
else puts("hopeless");
}
return ;
}