SGU 103 Traffic Lights 经典最短路
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103. Traffic Lights
Time limit per test: 0.5 second(s)
Memory limit: 4096 kilobytes
Memory limit: 4096 kilobytes
input: standard
output: standard
output: standard
In the city of Dingilville the traffic is arranged in an unusual way. There are junctions and roads connecting the junctions. There is at most one road between any two different junctions. There is no road connecting a junction to itself. Travel time for a road is the same for both directions. At every junction there is a single traffic light that is either blue or purple at any moment. The color of each light alternates periodically: blue for certain duration and then purple for another duration. Traffic is permitted to travel down the road between any two junctions, if and only if the lights at both junctions are the same color at the moment of departing from one junction for the other. If a vehicle arrives at a junction just at the moment the lights switch it must consider the new colors of lights. Vehicles are allowed to wait at the junctions. You are given the city map which shows:
Your task is to find a path which takes the minimum time from a given source junction to a given destination junction for a vehicle when the traffic starts. In case more than one such path exists you are required to report only one of them.
Input
The first line contains two numbers: The id-number of the source junction and the id-number of the destination junction. The second line contains two numbers: N, M. The following N lines contain information on N junctions. The (i+2)'th line of the input file holds information about the junction i : Ci, riC, tiB, tiP where Ci is either B for blue or P forpurple, indicating the initial color of the light at the junction i. Finally, the next M lines contain information on M roads. Each line is of the form: i, j, lij where i and j are the id-numbers of the junctions which are connected by this road. 2 ≤ N ≤ 300 where N is the number of junctions. The junctions are identified by integers 1 through N. These numbers are called id-numbers. 1 ≤ M ≤ 14000 where M is the number of roads. 1 ≤ lij ≤ 100 where lij is the time required to move from junction i to j using the road that connects i and j. 1 ≤ tiC ≤ 100 where tiC is the duration of the color c for the light at the junction i. The index c is either 'B' for blue or 'P' for purple. 1 ≤ riC ≤ tiC where riC is the remaining time for the initial color c at junction i.
Output
If a path exists:If a path does not exist:
Example(s)
sample input |
sample output |
1 4 4 5 B 2 16 99 P 6 32 13 P 2 87 4 P 38 96 49 1 2 4 1 3 40 2 3 75 2 4 76 3 4 77 |
127 1 2 4 |
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边权随着时间变化。可以证明dijkstra的贪心策略仍然适用。
所以只要计算出两点间的等待时间即可。
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int maxn=1111;;
const int maxm=41111;
const int INF=0x3f3f3f3f;
int st,ed;
int n,m;
struct Node{
char c[2];
int r,tb,tp;
}a[maxn];
class Dijkstra{
private:
struct EdgeNode{
int to,w,next;
};
struct HeapNode{
int u,d;
HeapNode(int _u,int _d):u(_u),d(_d){}
bool operator<(const HeapNode& rhs) const{
return d>rhs.d;
}
};
EdgeNode edges[maxm];
int head[maxn],edge;
int dis[maxn],pre[maxn];
bool vis[maxn];
priority_queue<HeapNode>que;
void color(int time,int x,int &px,char cx[]){
if (time<a[x].r){
px=a[x].r-time;
cx[0]=a[x].c[0];
return;
}
int t=(time-a[x].r)%(a[x].tb+a[x].tp);
if (a[x].c[0]=='B'&&t<a[x].tp){
px=a[x].tp-t;
cx[0]='P';
}
else if (a[x].c[0]=='P'&&t<a[x].tb){
px=a[x].tb-t;
cx[0]='B';
}
else{
px=a[x].tb+a[x].tp-t;
if (a[x].c[0]=='B') cx[0]='B';
else cx[0]='P';
}
}
public:
int getTime(int time,int x,int y,int num){
if (num>3) return -1;
int px,py;
char cx[2],cy[2];
color(time,x,px,cx);
color(time,y,py,cy);
if (cx[0]==cy[0]) return time;
if (px==py) return getTime(time+px,x,y,num+1);
time+=min(px,py);
return time;
}
void addedge(int u,int v,int w){
edges[edge].w=w,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}
void init(){
memset(head,-1,sizeof(head));
edge=0;
}
int getDis(int i){
return dis[i];
}
int getPre(int i){
return pre[i];
}
void dijkstra(int s){
while (!que.empty()) que.pop();
memset(dis,0x3f,sizeof(dis));
memset(vis,0,sizeof(vis));
memset(pre,-1,sizeof(pre));
dis[s]=0;
que.push(HeapNode(s,0));
while (!que.empty()){
HeapNode top=que.top();
que.pop();
int u=top.u;
if (vis[u]) continue;
vis[u]=true;
for (int i=head[u];i!=-1;i=edges[i].next){
int v=edges[i].to;
int w=edges[i].w;
int time=getTime(dis[u],u,v,1);
if (time==-1) continue;
if (dis[v]>time+w){
dis[v]=time+w;
pre[v]=u;
que.push(HeapNode(v,dis[v]));
}
}
}
}
void output(int u){
if (pre[u]!=-1) output(pre[u]);
else{
printf("%d",u);
return;
}
printf(" %d",u);
}
}dij;
void init(){
dij.init();
}
void input(){
scanf("%d%d",&st,&ed);
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++){
scanf("%s%d%d%d",a[i].c,&a[i].r,&a[i].tb,&a[i].tp);
}
for (int i=1;i<=m;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
dij.addedge(u,v,w);
dij.addedge(v,u,w);
}
}
void solve(){
dij.dijkstra(st);
if (dij.getDis(ed)==INF) printf("0\n");
else{
printf("%d\n",dij.getDis(ed));
dij.output(ed);
printf("\n");
}
}
int main()
{
init();
input();
solve();
return 0;
}
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