UVa

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<<1];
int p[302],tot;
int n,m,st,ed;
ll dis[MAX];

void dij(){
    bool f[302]={0};
    for(int i=1;i<=n;i++){
        dis[i]=INF;
    }
    dis[st]=0;
    for(int i=0;i<n;i++){
        int u;
        ll dd=INF;
        for(int j=1;j<=n;j++){
            if(!f[j] && dd>dis[j]) {
                dd=dis[j];
                u=j;
            }
        }
        if(dd==INF) break;
        f[u]=1;
        for(int j=p[u];j!=-1;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=1;
    while(scanf("%d %d %d %d",&n,&m,&st,&ed)!=EOF){
        memset(p,-1,sizeof(p));
        tot=0;
        for(int i=0;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
",t++,dis[ed]);

    }
    return 0;
}