uva 11374 最短路+记录路径 dijkstra最短路模板

UVA - 11374

Airport Express

Time Limit:1000MS

Memory Limit:Unknown

64bit IO Format:%lld & %llu

[Submit] [Go Back] [

id=22966" style="color:rgb(106,57,6); text-decoration:none">Status]

Description

ProblemD: Airport Express

In a small city called Iokh, a train service, Airport-Express, takes residents to the airport more quickly than other transports. There are two types of trains in Airport-Express, theEconomy-Xpress and theCommercial-Xpress.
They travel at different speeds, take different routes and have different costs.

Jason is going to the airport to meet his friend. He wants to take the Commercial-Xpress which is supposed to be faster, but he doesn't have enough money. Luckily he has a ticket for the Commercial-Xpress which can take him one station forward. If he used
the ticket wisely, he might end up saving a lot of time. However, choosing the best time to use the ticket is not easy for him.

Jason now seeks your help. The routes of the two types of trains are given. Please write a program to find the best route to the destination. The program should also tell when the ticket should be used.

Input

The input consists of several test cases. Consecutive cases are separated by a blank line.

The first line of each case contains 3 integers, namely N,S andE (2 ≤N ≤ 500, 1 ≤S,E ≤N),
which represent the number of stations, the starting point and where the airport is located respectively.

There is an integer M (1 ≤ M ≤ 1000) representing the number of connections between the stations of the Economy-Xpress. The nextM lines give the information of the routes of the
Economy-Xpress. Each consists of three integersX, Y and Z (X,Y ≤N, 1 ≤Z ≤
100). This meansX andY are connected and it takesZ minutes to travel between these two stations.

The next line is another integer K (1 ≤ K ≤ 1000) representing the number of connections between the stations of the Commercial-Xpress. The nextK lines contain the information
of the Commercial-Xpress in the same format as that of the Economy-Xpress.

All connections are bi-directional. You may assume that there is exactly one optimal route to the airport. There might be cases where you MUST use your ticket in order to reach the airport.

Output

For each case, you should first list the number of stations which Jason would visit in order. On the next line, output "TicketNot Used" if you decided NOT to use the ticket; otherwise, state the station where Jason should get on the train
of Commercial-Xpress. Finally, print thetotal time for the journey on the last line. Consecutive sets of output must be separated by a blank line.

Sample Input

Sample Output

Problemsetter: Raymond Chun

Originally appeared in CXPC, Feb. 2004

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=22966

由于仅仅能做一次商业线。我们能够枚举商业线T(a,b),则总时间为f(a)+T(a,b)+g(b);f和g用两次dijkstra来计算，以S为起点的dijkstra和以E为起点的dijkstra；

#include <iostream>

#include <cstdio>

#include <vector>

#include <cstring>

#include <algorithm>

#include <queue>

using namespace std;

const int MAXN = 505;

const int INF = 0x3f3f3f3f;

struct Edge {

int from, to, dist;

};

struct HeapNode {

int d, u;

bool operator< (const HeapNode rhs) const {

return d > rhs.d;

}

};

struct Dijkstra {

int n, m; // 点数和边数

vector<Edge> edges; //边列表

vector<int> G[MAXN]; // 每一个点出发的边编号（0開始）

bool done[MAXN]; // 是否已标记

int d[MAXN]; //s 到各个点的距离

int p[MAXN]; //最短路中上一个点，也能够是上一条边

void init(int n) {

this->n = n;

for (int i = 0; i < n; i++)

G[i].clear();

edges.clear();

}

void AddEdge(int from, int to, int dist) {

edges.push_back((Edge){from, to, dist});

m = edges.size();

G[from].push_back(m-1);

}

void dijkstra(int s) {

priority_queue<HeapNode> Q;

for (int i = 0; i < n; i++)

d[i] = INF;

d[s] = 0;

memset(done, 0, sizeof(done));

Q.push((HeapNode){0, s});

while (!Q.empty()) {

HeapNode x = Q.top();

Q.pop();

int u = x.u;

if (done[u])

continue;

done[u] = true;

for (int i = 0; i < G[u].size(); i++) {

Edge &e = edges[G[u][i]];

if (d[e.to] > d[u] + e.dist) {

d[e.to] = d[u] + e.dist;

p[e.to] = e.from;

Q.push((HeapNode){d[e.to], e.to});

}

}

}

}

void getPath(vector<int> &path, int s, int e) {

int cur = e;

while (1) {

path.push_back(cur);

if (cur == s)

return ;

cur = p[cur];

}

}

};

int n, m, k, s, e;

int x, y, z;

vector<int> path;

int main() {

int first = 1;

while (scanf("%d%d%d", &n, &s, &e) != EOF) {

if (first)

first = 0;

else printf("\n");

s--, e--;

Dijkstra ans[2];

ans[0].init(n);

ans[1].init(n);

scanf("%d", &m);

while (m--) {

scanf("%d%d%d", &x, &y, &z);

x--, y--;

ans[0].AddEdge(x, y, z);

ans[0].AddEdge(y, x, z);

ans[1].AddEdge(x, y, z);

ans[1].AddEdge(y, x, z);

}

ans[0].dijkstra(s);

ans[1].dijkstra(e);

scanf("%d", &k);

path.clear();

int Min = ans[0].d[e];

int flagx = -1, flagy = -1;

while (k--) {

scanf("%d%d%d", &x, &y, &z);

x--, y--;

if (Min > ans[0].d[x] + z + ans[1].d[y]) {

Min = ans[0].d[x] + z + ans[1].d[y];

flagx = x, flagy = y;

}

if (Min > ans[1].d[x] + z + ans[0].d[y]) {

Min = ans[1].d[x] + z + ans[0].d[y];

flagx = y, flagy = x;

}

}

if (flagx == -1) //推断是否须要坐商业线

{

ans[0].getPath(path, s, e);

reverse(path.begin(), path.end());

for (int i = 0; i < path.size()-1; i++)

printf("%d ", path[i]+1);

printf("%d\n", path[path.size()-1]+1);

printf("Ticket Not Used\n");

printf("%d\n", Min);

}

else {

ans[0].getPath(path, s, flagx);

reverse(path.begin(), path.end());

ans[1].getPath(path, e, flagy);

for (int i = 0; i < path.size()-1; i++)

printf("%d ", path[i]+1);

printf("%d\n", path[path.size()-1]+1);

printf("%d\n", flagx+1);

printf("%d\n", Min);

}

}

return 0;

}