sgu 103 Traffic Lights

题目描述：

103. Traffic Lights

Time limit per test: 0.5 second(s)
Memory limit: 4096 kilobytes

input: 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:

the travel times for all roads (integers)

the durations of the two colors at each junction (integers)

and the initial color of the light and the remaining time (integer) for this color to change at each junction.

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 followingN lines contain information on N junctions. The (i+2)'th line of the input file holds information about the junctioni :C_i,r_iC, t_iB,t_iP whereC_i is eitherB forblue or P forpurple, indicating the initial color of the light at the junctioni. Finally, the nextM lines contain information onM roads. Each line is of the form:i,j, l_ij wherei and j are the id-numbers of the junctions which are connected by this road. 2 ≤N ≤ 300 whereN is the number of junctions. The junctions are identified by integers 1 throughN. These numbers are called id-numbers. 1 ≤M ≤ 14000 whereM is the number of roads. 1 ≤ l_ij ≤ 100 wherel_ij is the time required to move from junctioni toj using the road that connectsi and j. 1 ≤t_iC ≤ 100 wheret_iC is the duration of the colorc for the light at the junctioni. The indexc is either 'B' for blue or 'P' for purple. 1 ≤r_iC ≤t_iC where r_iC is the remaining time for the initial colorc at junctioni.

Output

If a path exists:

The first line will contain the time taken by a minimum-time path from the source junction to the destination junction.

Second line will contain the list of junctions that construct the minimum-time path you have found. You have to write the junctions to the output file in the order of travelling. Therefore the first integer in this line must be the id-number of the source junction and the last one the id-number of the destination junction.

If a path does not exist:

A single line containing only the integer 0.

Example(s)

sample input

sample output

1 44 5B 2 16 99P 6 32 13P 2 87 4P 38 96 491 2 41 3 402 3 752 4 763 4 77

1271 2 4

这道题WA了无数次a......，后来才发现是将<=写成==了。。（RP啊）

验证了我的百折不挠。

好了，下面是思路：

是属于路联通性会变化的最短路，我们需要判断t时刻要从u到v我们需要等待的时间。

考虑到我们不会等待超过两个回合，我们可以判断一下，防止TLE(苦逼的TLE。。。)

然后就很好写了。我用的是spfa ，效果还可以。。

附上代码:

#include<iostream>#include<cstring>#include<cstdio>#include<set>#include<algorithm>#include<vector>#include<cstdlib>#include<queue>#define inf 0xfffffff#define CLR(a,b) memset((a),(b),sizeof((a)))using namespace std;int const nMax = 310;typedef int LL;typedef pair<LL,LL> pij;int dist[nMax],tt[nMax];int c[nMax],r[nMax],t[nMax][2];  //0->blue 1->purplevector<pij> Q[nMax];int n,m;int so,si;char s[nMax];int vis[nMax];queue<int> q;int fa[nMax];int inline f(int u,int T){    T%=(t[u][0]+t[u][1]);    if((T<r[u])||(T>=r[u]+t[u][c[u]^1]))return c[u];    return c[u]^1;}int inline next_t(int u,int T){    int _t=T%(t[u][0]+t[u][1]);    if((_t<r[u]))return T+r[u]-_t;    else if(_t>=r[u]&&_t<r[u]+t[u][c[u]^1])return T+r[u]+t[u][c[u]^1]-_t;    else return T+r[u]+t[u][0]+t[u][1]-_t;}int inline next(int u,int v,int T){    int cu=f(u,T),cv=f(v,T);    int tu=T;    int step=0;    while(cu!=cv){        tu=min(next_t(u,tu),next_t(v,tu));        cu=f(u,tu),cv=f(v,tu);        step++;        if(step>3)return -1;    }    return tu;}void inline output(int u){    if(fa[u]==-1){        printf("%d",u);        return ;    }    output(fa[u]);    printf(" %d",u);    return ;}void spfa(){    int u,v,w;    for(int i=1;i<=n;i++){        dist[i]=inf;        vis[i]=0;        tt[i]=0;    }    while(!q.empty())q.pop();    dist[so]=0;    q.push(so);    fa[so]=-1;    while(!q.empty()){        u=q.front();q.pop();        vis[u]=0;        for(int i=0;i<Q[u].size();i++){            v=Q[u][i].first,w=Q[u][i].second;            int add=next(u,v,dist[u]);            if(add==-1)continue;            if(dist[v]>w+add){                fa[v]=u;                dist[v]=w+add;                if(!vis[v]){                    q.push(v);                    vis[v]=1;                }            }         //   printf("u=%d v=%d add=%d dist[%d]=%d dist[%d]=%d\n",u,v,add,u,dist[u],v,dist[v]);        }    }    if(dist[si]==inf)printf("0\n");    else {        printf("%d\n",dist[si]);        output(si);    }}int main(){    int u,v,w;    scanf("%d%d",&so,&si);    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++){        scanf("%s %d %d %d",s,&r[i],&t[i][0],&t[i][1]);        c[i]=s[0]=='B'?0:1;        Q[i].clear();    }    for(int i=0;i<m;i++){        scanf("%d%d%d",&u,&v,&w);        Q[u].push_back(pij(v,w));        Q[v].push_back(pij(u,w));    }    spfa();    return 0;}