XYZZY uva BFS+DFS

Problem D: XYZZY

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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.

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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 thefinish.Each room has anenergy 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) ton (the finish room). Input for then 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

50 1 2-60 1 3-60 1 420 1 50 050 1 220 1 3-60 1 4-60 1 50 050 1 221 1 3-60 1 4-60 1 50 050 1 220 2 1 3-60 1 4-60 1 50 0-1

Output for Sample Input

hopelesshopelesswinnablewinnable

解决方案：首先，若是无环，直接DFS就能解决问题；当有环时，就判断，每走一次该环是否都比前一次的能量多，若是，则BFS看从该环出发能否到达终点。

代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int Map[120][120];bool vis[120];int E_room[120],energy[120],Q[120];//energy记录走到每个的room时的能量，E_room进入房间能获得的能量int head,tail;int n;bool bfs(int k){    memset(vis,false,sizeof(vis));    head=tail=0;    Q[head++]=k;    vis[k]=true;    while(head!=tail)    {        int temp=Q[tail++];        if(temp==n) return true;        for(int i=1;i<=n;i++)        {            if(!vis[i]&&Map[temp][i])            {                vis[i]=true;                Q[head++]=i;            }        }    }    return false;}bool dfs(int S,int E){    int sum=E;    if(S==n) return true;    for(int i=1;i<=n;i++)    {        if(Map[S][i]&&sum+E_room[i]>0)        {            if(!energy[i])            {                energy[i]=sum+E_room[i];                if(dfs(i,energy[i]))                    return true;            }            else if(energy[i]<sum+E_room[i]&&bfs(i))                return true;        }    }    return false;}int main(){     while(~scanf("%d",&n)&&(n!=-1))     {         memset(Map,0,sizeof(Map));         memset(energy,0,sizeof(energy));         for(int i=0;i<n;i++)         {             int m;             scanf("%d %d",&E_room[i+1],&m);             for(int j=0;j<m;j++)             {                 int temp;                 scanf("%d",&temp);                 Map[i+1][temp]=1;             }         }         if(dfs(1,100))            cout<<"winnable\n";         else cout<<"hopeless\n";     }    return 0;}