[UVA][SSSP] 11374 - Airport Express
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Problem D: 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, the Economy-Xpress and the Commercial-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 and E (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 next M lines give the information of the routes of the Economy-Xpress. Each consists of three integers X, Y and Z (X, Y ≤ N, 1 ≤ Z ≤ 100). This means X and Y are connected and it takes Z 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 next K 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 "Ticket Not 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 the total time for the journey on the last line. Consecutive sets of output must be separated by a blank line.
Sample Input
4 1 4 4 1 2 2 1 3 3 2 4 4 3 4 5 1 2 4 3
Sample Output
1 2 4 2 5
Problemsetter: Raymond Chun
Originally appeared in CXPC, Feb. 2004
題目描述：
要求從起點到終點的最短路，但是 Jason 有一次機會使用特殊道路，
而這條特殊道路有可能使最短路更短，求最短的最短路徑。
並且輸出路徑走法。
題目解法：
無向圖正權最短路，代碼中使用 SPFA 計算。
分別對起點和終點各做一次 SPFA，然後對於特殊道路做窮舉，即可在 O(1) 時間內找到變更的最短路徑。
輸出稍微麻煩了些。
#include <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
struct edge {
int to, v;
edge(int a, int b):
to(a), v(b) {}
};
vector<edge> g[505];
void spfa(int n, int st, int dist[], int prev[]) {
int i, j, k;
int tn, inq[505] = {};
queue<int> Q;
for(i = 1; i <= n; i++)
dist[i] = 0xfffffff;
dist[st] = 0, Q.push(st);
while(!Q.empty()) {
tn = Q.front(), Q.pop();
inq[tn] = 0;
for(i = 0; i < g[tn].size(); i++) {
if(dist[g[tn][i].to] > dist[tn] + g[tn][i].v) {
dist[g[tn][i].to] = dist[tn] + g[tn][i].v;
prev[g[tn][i].to] = tn;
if(inq[g[tn][i].to] == 0) {
inq[g[tn][i].to] = 1;
Q.push(g[tn][i].to);
}
}
}
}
}
int main() {
int n, m, st, ed;
int i, j, k, x, y, v;
int cases = 0;
while(scanf("%d %d %d", &n, &st, &ed) == 3) {
if(cases++) puts("");
for(i = 1; i <= n; i++)
g[i].clear();
scanf("%d", &m);
while(m--) {
scanf("%d %d %d", &x, &y, &v);
g[x].push_back(edge(y, v));
g[y].push_back(edge(x, v));
}
int distST[505], distED[505];
int prevST[505], prevED[505];
spfa(n, st, distST, prevST);
spfa(n, ed, distED, prevED);
int rx = -1, ry = -1, mn = distST[ed];
scanf("%d", &m);
while(m--) {
scanf("%d %d %d", &x, &y, &v);
if(distST[x] + distED[y] + v < mn) {
mn = distST[x] + distED[y] + v;
rx = x, ry = y;
}
if(distST[y] + distED[x] + v < mn) {
mn = distST[y] + distED[x] + v;
rx = y, ry = x;
}
}
int rrx = rx;
if(rx == -1) {
while(st != ed) {
printf("%d ", st);
st = prevED[st];
}
printf("%d\n", ed);
} else {
int buffer[505], bidx = 0;
while(rx != st) {
buffer[bidx++] = rx;
rx = prevST[rx];
}
buffer[bidx++] = st;
for(i = bidx-1; i >= 0; i--) {
printf("%d", buffer[i]);
if(i) putchar(' ');
}
while(ry != ed) {
printf(" %d", ry);
ry = prevED[ry];
}
printf(" %d\n", ed);
}
if(rrx == -1)
puts("Ticket Not Used");
else
printf("%d\n", rrx);
printf("%d\n", mn);
}
return 0;
}