UVa - 12661 - Funny Car Racing

先上题目：

12661 Funny Car Racing
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

　　题意：给出一个有向图，每条边都有三个值，打开时间间隔，关闭时间间隔，通过它需要多少时间，路的开关是循环往复的。给你起点和终点，问你最少需要多少时间才可以从起点到达终点(保证一定存在这种路)。

　　这其实是求最短路，用DIJ求最短路，然后对于从某个点开始可以到达的下一个点，在分析是否需要更新的时候，需要判断当前时刻路开了没有，如果开了看看够不够时间过去，然后根据这些情况更新最小值就可以了。这里需要注意的东西就是这事有向图不是无向图。

上代码：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #define MAX 50002
 #define ll long long
 #define INF (((ll)1)<<60)
 using namespace std;
 
 typedef struct edge{
     int to,next;
     ll a,b,c,s;
 }edge;
 
 edge e[MAX<<];
 int p[],tot;
 int n,m,st,ed;
 ll dis[MAX];
 
 void dij(){
     bool f[]={};
     for(int i=;i<=n;i++){
         dis[i]=INF;
     }
     dis[st]=;
     for(int i=;i<n;i++){
         int u;
         ll dd=INF;
         for(int j=;j<=n;j++){
             if(!f[j] && dd>dis[j]) {
                 dd=dis[j];
                 u=j;
             }
         }
         if(dd==INF) break;
         f[u]=;
         for(int j=p[u];j!=-;j=e[j].next){
             ll t=dis[u];
             ll r=t%e[j].s;
             if(r>=e[j].a){
                 t+=(e[j].s-r)+e[j].c;
             }else{
                 if(e[j].a-r>=e[j].c) t+=e[j].c;
                 else t+=e[j].s-r+e[j].c;
             }
             dis[e[j].to]=min(t,dis[e[j].to]);
         }
     }
 }
 
 inline void add(int u,int v,int a,int b,int c){
     e[tot].to=v;    e[tot].next=p[u];
     e[tot].a=a;     e[tot].b=b;     e[tot].c=c;     e[tot].s=a+b;
     p[u]=tot++;
 }
 
 int main()
 {
     int t,u,v,a,b,c;
     //freopen("data.txt","r",stdin);
     t=;
     while(scanf("%d %d %d %d",&n,&m,&st,&ed)!=EOF){
         memset(p,-,sizeof(p));
         tot=;
         for(int i=;i<m;i++){
             scanf("%d %d %d %d %d",&u,&v,&a,&b,&c);
             if(c<=a){
                 add(u,v,a,b,c);
                 //add(v,u,a,b,c);
             }
         }
         dij();
         printf("Case %d: %lld\n",t++,dis[ed]);
 
     }
     return ;
 }

/*12661*/