Funny Car Racing UVA - 12661

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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

最短路

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int MAX=1e5+100;
const int inf = 1<<29;
int e,n,m,s,t,head[MAX],nex[MAX],cost[MAX],a[MAX],b[MAX],pnt[MAX],dist[MAX];
bool vis[MAX];
queue<int> q;
void add(int u,int v,int c,int sa,int sb)
{
    pnt[e] = v;
    nex[e] = head[u];
    cost[e] = c;
    a[e] = sa;
    b[e] = sb;
    head[u] = e++;
}
int spfa(int st,int des)
{
    int i,j;
    for(i=0;i<=n;i++)
        dist[i] = inf;
    dist[st] = 0;
    memset(vis,0,sizeof(vis));
    q.push(st);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(i=head[u];i!=-1;i=nex[i])
        {
            int v = pnt[i];
            int val = dist[u],s= dist[u];
            val = val%(a[i]+b[i]);
            if(val > a[i])
                s += b[i] - (val-a[i]);
            else if(a[i]-val < cost[i])
                s+=b[i]+a[i]-val;
            if(s+cost[i]<dist[v])
            {
                dist[v] = s+cost[i];
                if(!vis[v])
                {
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }
    }
    return dist[des];
}
int main()
{
    int cas = 1;
    while(scanf("%d%d%d%d",&n,&m,&s,&t) != EOF)
    {
        e = 0;
        memset(head,-1,sizeof(head));
        for(int i=0;i<m;i++)
        {
            int u,v,a,b,t;
            scanf("%d%d%d%d%d",&u,&v,&a,&b,&t);
            if(a>=t)
                add(u,v,t,a,b);
        }
        printf("Case %d: %d\n",cas++,spfa(s,t));
    }
    return 0;
}