Funny Car Racing
Description
There is a funny car racing in a city with n junctions and m directed roads.
The funny part is: each road is open and closed periodically. Each road is associate with two integers (a, b), that means the road will be open for a seconds, then closed for b seconds, then open for a seconds… All these start from the beginning of the race. You must enter a road when it’s open, and leave it before it’s closed again.
Your goal is to drive from junction s and arrive at junction t as early as possible. Note that you can wait at a junction even if all its adjacent roads are closed.
Input
There will be at most 30 test cases. The first line of each case contains four integers n, m, s, t (1<=n<=300, 1<=m<=50,000, 1<=s,t<=n). Each of the next m lines contains five integers u, v, a, b, t (1<=u,v<=n, 1<=a,b,t<=105), that means there is a road starting from junction u ending with junction v. It’s open for a seconds, then closed for b seconds (and so on). The time needed to pass this road, by your car, is t. No road connects the same junction, but a pair of junctions could be connected by more than one road.
Output
For each test case, print the shortest time, in seconds. It’s always possible to arrive at t from s.
Sample Input
3 2 1 3
1 2 5 6 3
2 3 7 7 6
3 2 1 3
1 2 5 6 3
2 3 9 5 6
Sample Output
Case 1: 20
Case 2: 9
Source
湖南省第九届大学生计算机程序设计竞赛
思路:基本上是很裸的spfa,在松弛条件处改一下就好了。
有一个坑点就是,那条路的t有可能大于a。(具体详见代码)
代码:
#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define maxn 100000+10
#define maxv 1000+10
const int inf=0x3f3f3f3f;
typedef long long LL;
LL n,m,len;
struct node
{
LL v,a,b,t,next;
}G[maxn];
LL d[maxv],first[maxv],inq[maxv];
void spfa(LL st,LL ed)
{
for(LL i=1;i<=n;i++)
{
d[i]=100000000000000;
inq[i]=0;
}
d[st]=0,inq[st]=1;
queue<LL>q;
q.push(st);
while(!q.empty())
{
st=q.front();
q.pop();
inq[st]=0;
for(LL i=first[st];i!=-1;i=G[i].next)
{
LL v=G[i].v,a=G[i].a,b=G[i].b,t=G[i].t;
LL cnt=(d[st]%(a+b)),tot;
if(t>a)
continue;
if(cnt<=a)
{
if((cnt+t)<=a)
tot=d[st]+t;
else
tot=a-cnt+b+t+d[st];
}
else
tot=d[st]+a+b-cnt+t;
if(d[v]>tot)
{
d[v]=tot;
if(!inq[v])
{
inq[v]=1;
q.push(v);
}
}
}
}
printf("%lld\n",d[ed]);
}
void add_egde(LL u,LL v,LL a,LL b,LL t)
{
G[len].v=v,G[len].a=a,G[len].b=b,G[len].t=t;
G[len].next=first[u];
first[u]=len++;
}
int main()
{
LL T=1,st,ed;
while(~scanf("%lld%lld%lld%lld",&n,&m,&st,&ed))
{
len=0;
mem(first,-1);
LL u,v,a,b,t;
for(LL i=0;i<m;i++)
{
scanf("%lld%lld%lld%lld%lld",&u,&v,&a,&b,&t);
add_egde(u,v,a,b,t);
}
printf("Case %lld: ",T++);
spfa(st,ed);
}
return 0;
}