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*/

巴特西

UVa - 12661 - Funny Car Racing

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