poj 1613 Cave Raider

Cave Raider

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 551		Accepted: 171

Description

Afkiyia is a big mountain. Inside the mountain, there are many caves. These caves are connected by tunnels. Hidden in one of the caves is a terrorist leader. Each tunnel connects two caves. There could be more than one tunnels connect the same two caves. 
At the joint of a tunnel and a cave, there is a door. From time to 
time, the terrorists close a tunnel by shutting the two doors at the two
 ends, and "clean" the tunnel. It is still a mystery how they clean the 
tunnel. However, we know that if a person (or any living creature) is 
trapped in the tunnel when it is being cleaned, then the person (or the 
living creature) will die. After a cleaning of the tunnel is finished, 
the door will open, and the tunnel can be used again.

Now the intelligence servicemen have found out which cave the leader
 is hiding,and moreover, they know the schedule of the cleaning of the 
tunnels. Jing Raider is going to go into the cave and catch the leader. 
You need to help him find a route so that he can get to that cave in the
 shortest time. Be careful not to be trapped in a tunnel when it is 
being cleaned.

Input

The
 input consists of a number of test cases. The 1st line of a test case 
contains four positive integers n,m, s, t, separated by at least one 
space, where n is the number of caves (numbered 1, 2, ... , n), m is the
 number of tunnels (numbered 1, 2, ... ,m), s is the cave where Jing is 
located at time 0, and t is the cave where the terrorist leader is 
hiding. (1 <= s, t <= n <= 50 and m <= 500).

The next m lines are information of the m tunnels: Each line is a 
sequence of at most 35 integers separated by at least one space. The 
first two integers are the caves that are the ends of the corresponding 
tunnel. The third integer is the time needed to travel from one end of 
the tunnel to the other. This is followed by an increasing sequence of 
positive integers (each integer is at most 10000) which are alternately 
the closing and the opening times of the tunnel. For example, if the 
line is

10 14 5 6 7 8 9

then it means that the tunnel connects cave 10 and cave 14, it takes
 5 units of time to go from one end to the other. The tunnel is closed 
at time 6, opened at time 7, then closed again at time 8, opened again 
at time 9. Note that the tunnel is being cleaned from time 6 to time 7, 
and then cleaned again from time 8 to time 9. After time 9, it remains 
open forever.

If the line is

10 9 15 8 18 23

then it means that the tunnel connects cave 10 and cave 9, it takes 
15 units of time to go from one end to the other. The tunnel is closed 
at time 8, opened at time 18,then closed again at time 23. After time 
23, it remains closed forever.

The next test case starts after the last line of the previous case. A 0 signals the end of the input.

Output

The output contains one line for each test case. Each line contains either an integer, which is the time needed for Jing to get to cave t, or the symbol *, which means that Jing can never get to cave t. Note that the starting time is 0. So if s = t, i.e., Jing is at the same cave as the terrorist leader, then the output is 0.

Sample Input

2 2 1 2
1 2 5 4 10 14 20 24 30
1 2 6 2 10 22 30
6 9 1 6
1 2 6 5 10
1 3 7 8 20 30 40
2 4 8 5 13 21 30
3 5 10 16 25 34 45
2 5 9 22 32 40 50
3 4 15 2 8 24 34
4 6 10 32 45 56 65
5 6 3 2 5 10 15
2 3 5 2 9 19 25
2 2 1 2
1 2 7 6 9 12
1 2 9 8 12 19
0

Sample Output

16
55
*

Source

Asia Kaohsiung 2003

sfpa 算法，
知道怎么更新就好了
队列的点出队列后，判断它能否更新临近的点
就是在隧道开开启的时间段他能否有时间通过
1A 比较开心的
#include <iostream>
#include <stdio.h>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 52
#define Max 1000000000
structnode
{
  int to;
  int cost;
  int next;
  int n;
  int t[43];
}Eg[1002];
int v[N];
char ch[1002];
int n,m,s,t;
int nu;
int rc [N];
void del(char *ch)
{
    int i;
    int id=0;
    int t=0;
    for(i=0;ch[i]!='\0';i++)
     if(ch[i]>='0'&&ch[i]<='9')
       t=t*10+ch[i]-'0';
     else if(t)
     {
          rc[id++]=t;
          t=0;
      }
    if(t) rc[id++]=t;
   int a=rc[0],b=rc[1],c=rc[2];
    Eg[nu].to=b;
    Eg[nu].n=id-2;
    Eg[nu].cost=c;
    Eg[nu].next=v[a];
    v[a]=nu++;
    Eg[nu].to=a;
    Eg[nu].n=id-2;
    Eg[nu].cost=c;
    Eg[nu].next=v[b];
    v[b]=nu;
    a=1;
    for(i=3;i<id;i++)
     {
         Eg[nu-1].t[a]=rc[i];
         Eg[nu].t[a++]=rc[i];
     }
     Eg[nu-1].t[a]=Max;//自加了个Max代表无穷远了
     Eg[nu].t[a++]=Max;
     nu++;
}
int dis[N];
int main()
{
    int i,k,a;
    while(scanf("%d %d %d %d",&n,&m,&s,&t)==4&&n)
    {
        memset(v,-1,sizeof(v));
        nu=0;gets(ch);
        while(m--)
        {
            gets(ch);
            del(ch);
        }
     if(s==t) { printf("0\n");continue;}
      for(i=1;i<=n;i++)
       dis[i]=Max;
      dis[s]=0;
      memset(rc,0,sizeof(rc));
      queue<int>Q;
      Q.push(s);rc[s]=1;
      int time,len;
      while(!Q.empty())
      {
          a=Q.front();Q.pop();
          rc[a]=0;
          for(i=v[a];i!=-1;i=Eg[i].next)
          {
              for(k=1;k<=Eg[i].n;k++)
                if(k%2)
                {
                   len=Eg[i].t[k]-Eg[i].t[k-1];
                   time=max(dis[a],Eg[i].t[k-1])+Eg[i].cost;
                   if(len>=Eg[i].cost&&Eg[i].t[k]>=time&&time<dis[Eg[i].to])//可以更新的条件
                   {
                      dis[Eg[i].to]=time;
                      if(!rc[Eg[i].to]) Q.push(Eg[i].to);
                      break;
                   }
                }
          }
      }
      if(dis[t]==Max) printf("*\n"); else printf("%d\n",dis[t]);

    }
    return 0;
}