Uva 省赛傻逼题 F

哎，现场一道煞笔题又没有出！

Problem F

 Time limit: 1.000 seconds

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

Output for the Sample Input

Case 1: 20
Case 2: 9

---

The Ninth Hunan Collegiate Programming Contest (2013)
Problemsetter: Rujia Liu
Special Thanks: Md. Mahbubul Hasan

 

#include <map>
#include <queue>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 666
#define ll long long
#define INF 0x7fffffff
struct Edge{int from,to,a,b,t;};
vector<int>G[maxn];
vector<Edge>edges;
int n,m;
int d[maxn];
int in[maxn];
void AddEdge(int from,int to,int a,int b,int t){
    edges.push_back((Edge){from,to,a,b,t});
    int m=edges.size();
    G[from].push_back(m-1);
}
void init(){
    for(int i=0;i<n;i++)G[i].clear();
    edges.clear();
}
int SPFA(int s){
    queue<int>q;
    for(int i=0;i<n;i++)d[i]=INF;
    d[s]=0;in[s]=1;q.push(s);
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=0;i<G[u].size();i++){
            Edge e=edges[G[u][i]];
            int wt;
            if(e.a-d[u]%(e.a+e.b)>=e.t)wt=0;
            else wt=e.a+e.b-d[u]%(e.a+e.b);
            if(e.a<e.t){
                if(d[u]%(e.a+e.b)<=e.a){
                    if((e.t+d[u]%(e.a+e.b))%(e.a+e.b)<=e.a)wt=0;
                    else wt=e.b-(e.t+d[u]%(e.a+e.b))%(e.a+e.b);
                }
                else {
                    if(e.t%(e.a+e.b)<=e.a);
                    else wt+=e.b-e.t%(e.a+e.b);
                }
            }
            if(d[e.to]>d[u]+e.t+wt){
                d[e.to]=d[u]+e.t+wt;
                if(!in[e.to]){
                    in[e.to]=1;
                    q.push(e.to);
                }
            }
        }
        in[u]=0;
    }
}
int main(){
    int s,e;
    int cas=1;
    while(scanf("%d%d%d%d",&n,&m,&s,&e)!=EOF){
        init();
        for(int i=0;i<m;i++){
            int u,v,a,b,t;
            scanf("%d%d%d%d%d",&u,&v,&a,&b,&t);
            AddEdge(u-1,v-1,a,b,t);
            //AddEdge(v-1,u-1,a,b,t);
        }
        printf("Case %d: ",cas++);
        SPFA(s-1);
        printf("%d\n",d[e-1]);
    }
    return 0;
}