UVALive

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
",path[path.size()-1]);
		}
	}
	return 0;
}