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
题意：
从房间1开始有100 energy value，每到新房间加上其ev，若为负则hopeless，能到房间n则winnable；
解题思路：
若到n一直都>0则winnable；
若到n路上有正环则winnable；
此外hopeless；
用Bellman_ford 判环
代码：
#include<bits/stdc++.h>
#define maxn 105
using namespace std;
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
return x*f;
}
int vis[maxn][maxn],d[maxn],n;
vector<pair<int,int> >e[maxn];
void floyd(){ //连通？
for(int k=1;k<=n;++k)
for(int i=1;i<=n;++i)
for(int j=1;j<=n;++j)
vis[i][j]=vis[i][j]||(vis[i][k]&&vis[k][j]);
}
int bellman_ford(int s,int t){
for(int i=1;i<=n;++i)d[i]=-1e9;
d[s]=100;
for(int k=0;k<n;++k)
for(int i=1;i<=n;i++)
for(int j=0;j<e[i].size();++j){
int v=e[i][j].first;
if(d[i]+e[i][j].second>0)
d[v]=max(d[v],d[i]+e[i][j].second);
}
for(int i=1;i<=n;i++)
for(int j=0;j<e[i].size();++j){
int v=e[i][j].first;
if(d[i]+e[i][j].second>0&&d[v]<d[i]+e[i][j].second&&vis[v][t]) return 1;
}
return d[t]>0;
}
int main(){
while((n=read())!=-1){
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;++i){
int val=read(),m=read();
while(m--){
int v=read();
e[i].push_back(make_pair(v,val));
vis[i][v]=1;
}
}
floyd();
if(bellman_ford(1,n))puts("winnable");
else puts("hopeless");
for(int i=1;i<=n;++i)e[i].clear();
}
}