XYZZY - UVa 10557 图论

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 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 thefinish. 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.

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.

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

题意：一个人一开始在1点有100能量，每次可以会得到或失去一定的能量，问是否可以从1走到n。

思路：这个考虑到有正环的原因。先预处理那些点可以走到n，如果一个可以走到n的点处于正环内，那么就一定可以走到n点。

AC代码如下：

#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<algorithm>using namespace std;int n,t,val[10010],f[10010],len[10010],d[10010];vector<int> vc1[10010],vc2[10010];void init(int v){    f[v]=1;    int i,u,length=vc2[v].size();    for(i=0;i<length;i++)    {        u=vc2[v][i];        if(f[u]!=1)           init(u);    }}bool dfs(int u,int e){    if(u==n)      return true;    int i,j,k,v;    for(i=0;i<len[u];i++)    {        v=vc1[u][i];        if(e+val[v]>0)        {            if(!d[v])            {                d[v]=e+val[v];                if(dfs(v,d[v]))                  return true;            }            else if(e+val[v]>d[v] && f[v]==1)              return true;        }    }    return false;}int main(){    int i,j,k,p;    while(~scanf("%d",&n) && n!=-1)    {        memset(f,0,sizeof(f));        t++;        for(i=1;i<=n;i++)        {            vc1[i].clear();            vc2[i].clear();        }        for(i=1;i<=n;i++)        {            scanf("%d",&val[i]);            scanf("%d",&k);            for(j=1;j<=k;j++)            {                scanf("%d",&p);                vc1[i].push_back(p);                vc2[p].push_back(i);            }        }        if(n<=1)        {            printf("winnable\n");            continue;        }        for(i=1;i<=n;i++)           len[i]=vc1[i].size();        init(n);        memset(d,0,sizeof(d));        if(dfs(1,100))          printf("winnable\n");        else          printf("hopeless\n");    }}