Funny Car Racing bfs题

D - 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：题意给你n个点m条边和起点位置和终点位置，然后开放的时间，关闭的时间，每个点有一个通过的时间，只有在开放时间才能过那个点，如果要过那个点的通过的时间大于那个点的开放的时间，那么这条边，就没用。

我用的是bfs搜索过的，我这里采用给每条边，编号，判重；

代码：

#include<stdio.h>#include<string.h>#include<queue>#include<vector>using namespace std;#define INF 999999999;struct edge{    int from,to;    int lu;    int red;    int w;    int id;};struct node{    int x;   long long  int step;    int father;    friend bool operator < (node n1,node n2)    {        return n1.step>n2.step; //<为从大到小，>大于从小到大；    }};vector<edge>edges;vector<int>G[301];bool vis[50001*2];int n,m,start,end1;int ans;int tot;void addedge(int x,int y,int lu,int red,int w){    edge a={x,y,lu,red,w,tot++};    edges.push_back(a);    G[x].push_back(edges.size()-1);}void bfs(){    memset(vis,0,sizeof(vis));    priority_queue<node>q;    node a={start,0,0};    q.push(a);    vis[0]=1;    while(!q.empty())    {        a=q.top();        q.pop();        if(a.x==end1)        {           if(ans>a.step) ans=a.step;           continue;        }        if(a.step>ans) continue;        for(int i=0;i<G[a.x].size();i++)        {            edge b=edges[G[a.x][i]];            if(vis[b.id]) continue;            vis[b.id]=1;            int kk=a.step%(b.red+b.lu);            int xx;            if(kk>b.lu)            {   kk=kk-b.lu;                xx=a.step+(b.red-kk)+b.w;                node bb={b.to,xx,a.x};                q.push(bb);            }            else            {                if(b.w<=b.lu-kk)                {                    node bb={b.to,a.step+b.w,a.x};                    q.push(bb);                }                else                {                    node bb={b.to,a.step+(b.lu-kk)+b.red+b.w,a.x};                    q.push(bb);                }            }        }    }}int main(){ int sss=1;  while(scanf("%d %d %d %d",&n,&m,&start,&end1)!=EOF)  {      for(int i=1;i<=n;i++)      {          G[i].clear();      }      edges.clear();      tot=1;      for(int i=1;i<=m;i++)      {          int x,y,lu,red,w;          scanf("%d %d %d %d %d",&x,&y,&lu,&red,&w);          if(w>lu) continue;          addedge(x,y,lu,red,w);      }     /* for(int i=0;i<edges.size();i++)      {          printf("%d ",edges[i].id);      }      printf("\n");*/      ans=999999999;      bfs();      printf("Case %d: %d\n",sss++,ans);  }}