题目链接：
http://acm.hust.edu.cn/vjudge/problem/48416
Shopping Malls
Time Limit: 3000MS
#### 问题描述
> We want to create a smartphone application to help visitors of a shopping mall and you have to calculate
> the shortest path between pairs of locations in the mall. Given the current location of the visitor and his
> destination, the application will show the shortest walking path (in meters) to arrive to the destination.
> The mall has N places in several floors connected by walking paths, lifts, stairs and escalators
> (automated stairs). Note that the shortest path in meters may involve using an escalator in the
> opposite direction. We only want to count the distance that the visitor has walked so each type of
> movement between places has a different cost in meters:
> • If walking or taking the stairs the distance is the euclidean distance between the points.
> • Using the lift has a cost of 1 meter because once we enter the lift we do not walk at all. One
> lift can only connect 2 points. An actual lift connects the same point of different floors, in the
> map all the points connected by a lift have the corresponding edge. So you do not need to worry
> about that. For instance, if there are three floors and one lift at position (1,2) of each floor, the
> input contains the edges (0, 1, 2) → (1, 1, 2), (1, 1, 2) → (2, 1, 2) and (0, 1, 2) → (2, 1, 2). In some
> maps it can be possible that a lift does not connect all the floors, then some of the edges will not
> be in the input.
> • The escalator has two uses:
> – Moving from A to B (proper direction) the cost is 1 meter because we only walk a few steps
> and then the escalator moves us.
> – Moving from B to A (opposite direction) has a cost of the euclidean distance between B and
> A multiplied by a factor of 3.
> The shortest walking path must use only these connections. All the places are connected to each
> other by at least one path.
#### 输入
> The input file contains several data sets, each of them as described below. Consecutive
> data sets are separated by a single blank line.
> Each data set contains the map of a unique shopping mall and a list of queries.
> The first line contains two integers N (N ≤ 200) and M (N −1 ≤ M ≤ 1000), the number of places
> and connections respectively. The places are numbered from 0 to N −1. The next N lines contain floor
> and the coordinates x, y of the places, one place per line. The distance between floors is 5 meters. The
> other two coordinates x and y are expressed in meters.
> The next M lines contain the direct connections between places. Each connection is defined by the
> identifier of both places and the type of movement (one of the following: ‘walking’, ‘stairs’, ‘lift’,
> or ‘escalator’). Check the cost of each type in the description above. The type for places in the same
> floor is walking.
> The next line contains an integer Q (1 ≤ Q ≤ 1000) that represents the number of queries that
> follow. The next Q lines contain two places each a and b. We want the shortest walking path distance
> to go from a to b.
#### 输出
> For each data set in the input the output must follow the description below. The outputs
> of two consecutive data sets will be separated by a blank line.
> For each query write a line with the shortest path in walked meters from the origin to the destination,
> with each place separated by a space.
样例
sample input
6 7
3 2 3
3 5 3
2 2 3
2 6 4
1 1 3
1 4 2
0 1 walking
0 2 lift
1 2 stairs
2 3 walking
3 4 escalator
5 3 escalator
4 5 walking
5
0 1
1 2
3 5
5 3
5 1
sample output
0 1
1 0 2
3 4 5
5 3
5 3 2 0 1
题解
floyd最短路+路径记录。注意重边和输出格式
代码
#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
using namespace std;
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;
const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const int maxn=222;
struct Point{
double x,y,z;
}pt[maxn];
double mat[maxn][maxn];
int pre[maxn][maxn];
int n,m;
double dis(const Point& p1,const Point& p2){
return sqrt(25*(p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)+(p1.z-p2.z)*(p1.z-p2.z));
}
void init(){
rep(i,0,n+1) rep(j,0,n+1){
mat[i][j]=INF*1.0;
if(i==j) mat[i][j]=0;
pre[i][j]=i;
}
}
int main(){
int kase=0;
while(scanf("%d%d",&n,&m)==2&&n){
init();
if(kase++) puts("");
rep(i,0,n) scanf("%lf%lf%lf",&pt[i].x,&pt[i].y,&pt[i].z);
while(m--){
int u,v; char str[22];
scanf("%d%d%s",&u,&v,str);
double val;
if(str[0]=='w'||str[0]=='s'){
mat[u][v]=min(mat[u][v],dis(pt[u],pt[v]));
mat[v][u]=min(mat[v][u],dis(pt[v],pt[u]));
}else if(str[0]=='e'){
mat[u][v]=min(mat[u][v],1.0);
mat[v][u]=min(mat[v][u],3*dis(pt[u],pt[v]));
}else{
mat[u][v]=min(mat[u][v],1.0);
mat[v][u]=min(mat[v][u],1.0);
}
}
rep(k,0,n) rep(i,0,n) rep(j,0,n){
if(mat[i][j]>mat[i][k]+mat[k][j]){
mat[i][j]=mat[i][k]+mat[k][j];
pre[i][j]=pre[k][j];
}
}
int q;
scanf("%d",&q);
while(q--){
int u,v;
scanf("%d%d",&u,&v);
VI path;
path.pb(v);
while(pre[u][v]!=u){
// bug(v);
path.pb(pre[u][v]);
v=pre[u][v];
}
path.pb(u);
reverse(all(path));
rep(i,0,path.size()-1) printf("%d ",path[i]);
printf("%d\n",path[path.size()-1]);
}
}
return 0;
}
