HDU 1317 XYZZY

http://acm.hdu.edu.cn/showproblem.php?pid=1317

XYZZY

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3017    Accepted Submission(s): 824

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
    

 

Source

University of Waterloo Local Contest 2003.09.27

 

这道题是做了很久，后来还是翻了别人的代码才做出来的。题意是，有n个房间，每个房间有能量值，（能量值有正有负），初始的时候你有100能量值，每到一个房间，你就会获得这个房间的能量值（负的就减少）.当你的能量值<0时你就输了！。问你能否走到第n个房间。如果走到就输出win 否则输出hopless。

解题思路：首先是如果这个图中有正环，则一定赢。如果不存在正环，那么就判断到 n房间时的能量值是否为正的。！

代码如下：flord判断联通。bellman-ford找最长路和判断正环。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>

#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF -1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b){
    return a>b;
}
const int N=103;
const int M=10004;
int energy[N];
int dp[N][N];
int dist[N];
int top;
///int edge[N][N];
struct node
{
    int s,e;
}edge[M];
int n;
void floyd()
{
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
          dp[i][j]=dp[i][j]||(dp[i][k]&&dp[k][j]);///进行判断是否联通
}
void bellman_ford()
{
    for(int i=1;i<=n;i++)///为什么是n。。这还不会？
        dist[i]=INF;
        dist[1]=100;
        for(int i=1;i<n;i++)
            for(int j=1;j<=top;j++)
            {
                int st=edge[j].s;
                int en=edge[j].e;
                if(dist[en]<dist[st]+energy[en]&&dist[st]+energy[en]>0)
                dist[en]=dist[st]+energy[en];
            }
               int mark=0;
               for(int j=1;j<=top;j++)///判断正环
               {
                  int st=edge[j].s;
                  int en=edge[j].e;
                  if(dist[en]<dist[st]+energy[en]&&dist[st]+energy[en]>0&&dp[en][n])///存在正环且能联通到n
                  {
                    mark=1;
                    cout<<"winnable"<<endl;
                    break;
                  }
               }
        if(!mark)
        {
            if(dist[n]>0)cout<<"winnable"<<endl;
            else cout<<"hopeless"<<endl;
        }
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int num,to;
    while(cin>>n)
    {
        top=0;
        cle(dp);
        for(int i=1;i<=n;i++)dp[i][i]=1;
        if(n==-1)break;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&energy[i],&num);
            for(int j=1;j<=num;j++)
            {
                scanf("%d",&to);
                top++;
                dp[i][to]=1;
                edge[top].s=i;
                edge[top].e=to;
            }
        }
        floyd();
        bellman_ford();
    }
    return 0;
}