uva 11374 最短路+記錄路徑 dijkstra最短路模板

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UVA - 11374

Airport Express

Time Limit:1000MS

Memory Limit:Unknown

64bit IO Format:%lld & %llu

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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

4 1 441 2 21 3 32 4 43 4 512 4 3

Sample Output

1 2 425

Problemsetter: Raymond Chun
Originally appeared in CXPC, Feb. 2004

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

因為只能做一次商業線，我們可以枚舉商業線T(a,b),則總時間為f(a)+T(a,b)+g(b);f和g用兩次dijkstra來計算，以S為起點的dijkstra和以E為起點的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;    // 點數和邊數
vector<Edge> edges;   //邊列表
vector<int> G[MAXN];  // 每個點出發的邊編號（0開始）
bool done[MAXN];   // 是否已標記
int d[MAXN];      //s 到各個點的距離
int p[MAXN]; //最短路中上一個點，也可以是上一條邊


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) //判斷是否需要坐商業線

                {
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;
}