CSU 1333 Funny Car Racing（spfa）

最新推荐文章于 2017-08-04 16:52:48 发布

浪漫些许潦草

最新推荐文章于 2017-08-04 16:52:48 发布

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分类专栏： # 最短路径 ACM 文章标签： Funny-Car Racing CSU-1333

本文链接：https://blog.youkuaiyun.com/blessLZH0108/article/details/69220107

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本文介绍了一种解决FunnyCar赛中寻找从起点到终点最短时间路径的问题。该问题涉及城市中的多个交叉口和定时开关的道路。通过修改SPFA算法来适应这种特殊的道路开关模式，确保赛车能在合适的时间通过每条道路。

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Funny Car Racing

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 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

Source
湖南省第九届大学生计算机程序设计竞赛

思路：基本上是很裸的spfa，在松弛条件处改一下就好了。

有一个坑点就是，那条路的t有可能大于a。（具体详见代码）

代码：

#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define maxn 100000+10
#define maxv 1000+10
const int inf=0x3f3f3f3f;
typedef long long  LL;
LL n,m,len;
struct node
{
    LL v,a,b,t,next;
}G[maxn];
LL d[maxv],first[maxv],inq[maxv];

void spfa(LL st,LL ed)
{
    for(LL i=1;i<=n;i++)
    {
        d[i]=100000000000000;
        inq[i]=0;
    }
    d[st]=0,inq[st]=1;
    queue<LL>q;
    q.push(st);
    while(!q.empty())
    {
        st=q.front();
        q.pop();
        inq[st]=0;
        for(LL i=first[st];i!=-1;i=G[i].next)
        {
            LL v=G[i].v,a=G[i].a,b=G[i].b,t=G[i].t;
            LL cnt=(d[st]%(a+b)),tot;
            if(t>a)
                continue;
            if(cnt<=a)
            {
                if((cnt+t)<=a)
                    tot=d[st]+t;
                else
                    tot=a-cnt+b+t+d[st];
            }
            else
                tot=d[st]+a+b-cnt+t;
            if(d[v]>tot)
            {
                d[v]=tot;
                if(!inq[v])
                {
                    inq[v]=1;
                    q.push(v);
                }
            }
        }
    }
    printf("%lld\n",d[ed]);
}

void add_egde(LL u,LL v,LL a,LL b,LL t)
{
    G[len].v=v,G[len].a=a,G[len].b=b,G[len].t=t;
    G[len].next=first[u];
    first[u]=len++;
}

int main()
{
    LL T=1,st,ed;
    while(~scanf("%lld%lld%lld%lld",&n,&m,&st,&ed))
    {
        len=0;
        mem(first,-1);
        LL u,v,a,b,t;
        for(LL i=0;i<m;i++)
        {
            scanf("%lld%lld%lld%lld%lld",&u,&v,&a,&b,&t);
            add_egde(u,v,a,b,t);
        }
        printf("Case %lld: ",T++);
        spfa(st,ed);
    }
    return 0;
}