1333: Funny Car Racing
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 482 Solved: 116
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 31 2 5 6 32 3 7 7 63 2 1 31 2 5 6 32 3 9 5 6
Sample Output
Case 1: 20Case 2: 9
HINT
Source
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define ll long long
const ll INF = 1LL<<60 ;
const ll N = 305 ;
struct EDG
{
ll to , next ;
ll a , b , c , d ;
} edg[50010];
ll eid , head[N] ;
ll dis[N] ;
void init()
{
eid = 0;
memset(head , -1 , sizeof(head));
for(ll i = 0 ; i < N ; i++)
dis[i] = INF ;
}
void addEdg(ll u , ll v , ll a , ll b , ll c)
{
edg[eid].to = v ;
edg[eid].next = head[u] ;
edg[eid].a = a ;
edg[eid].b = b ;
edg[eid].c = c ;
edg[eid].d = a + b ;
head[u] = eid++;
}
void spfa(ll s , ll t)
{
queue<ll>q;
bool inq[N] = { 0 } ;
ll u , v ;
dis[s] = 0 ;
if(s != t )
q.push( s ) ;
while(!q.empty())
{
u = q.front() ;
q.pop() ;
inq[u] = 0 ;
for(ll i = head[u] ; i!=-1; i=edg[i].next)
{
v = edg[i].to ;
ll tt = edg[i].a - (dis[u]%edg[i].d);
if(tt<edg[i].c)
tt += edg[i].b ;
else
tt = 0 ;
if(dis[v] > dis[u] + tt +edg[i].c)
{
dis[v] = dis[u] + tt +edg[i].c ;
if(inq[v]==0&&v!=t)
inq[v] = 1 , q.push(v);
}
}
}
}
int main()
{
ll n , m , s , t , u , v , a , b , c ;
ll T = 0;
while(scanf("%lld%lld%lld%lld",&n,&m,&s,&t)>0)
{
init();
while(m--)
{
scanf("%lld%lld%lld%lld%lld",&u , &v , &a , &b , &c) ;
if(c<=a)
addEdg( u , v , a , b , c );
}
spfa(s , t ) ;
if(dis[t]==INF)
dis[t] = -1;
printf("Case %lld: %lld\n",++T,dis[t]);
}
}
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