UVa 11374 Airport Express

题目大意：

给你n个点和起点终点，再告诉你m条经济线路和k条商业线路，经济线路可以无限制经过，但商业线路只能乘坐一次，问从起点到达终点的最短距离和路径，并输出在什么位置换成商业线路。

分析：

枚举每一条商业线路， 计算在当前线路换乘的最小费用， 并更新距离即可，注意在这之前要预处理两次，分别从起点和终点出发计算距离和路径。

代码：

#include<bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 500 + 10;

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;
vector<edge> edges;
vector<int> G[maxn];
bool done[maxn];
int dis[maxn], pre[maxn];

void init(int n) {
this -> n = n;
for(int i=0; i<n; i++) G[i].clear();
edges.clear();
}
void add_edge(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++) dis[i] = inf;
dis[s] = 0;
memset(pre, 0, sizeof(pre));
memset(done, false, 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<(int)G[u].size(); i++) {
edge& e = edges[G[u][i]];
if(dis[u] + e.dist < dis[e.to]) {
dis[e.to] = dis[u] + e.dist;
pre[e.to] = G[u][i];
Q.push((Heapnode){dis[e.to], e.to});
}
}
}
}
void Getshortestpaths(int s, int *dist, vector<int>* paths) {
dijkstra(s);
for(int i=0; i<n; i++) {
dist[i] = dis[i];
paths[i].clear();
int t = i;
paths[i].push_back(t);
while(t != s) {
paths[i].push_back(edges[pre[t]].from);
t = edges[pre[t]].from;
}
reverse(paths[i].begin(), paths[i].end());
}
}

}dij;

int u, v, w;
int n, m, k, s, t, Case;
int dis1[maxn], dis2[maxn];
vector<int> paths1[maxn], paths2[maxn];

void init() {
s--, t--;
dij.init(n);
scanf("%d", &m);
for(int i=0; i<m; i++) {
scanf("%d%d%d", &u, &v, &w); u--, v--;
dij.add_edge(u, v, w);
dij.add_edge(v, u, w);
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("data.txt", "r", stdin);
freopen("ans.txt", "w", stdout);
#endif
while(scanf("%d%d%d", &n, &s, &t) == 3) {
init();

dij.Getshortestpaths(s, dis1, paths1);
dij.Getshortestpaths(t, dis2, paths2);

int Ticket = -1, Time = dis1[t];
vector<int> path = paths1[t];

scanf("%d", &k);
for(int i=0; i<k; i++) {
scanf("%d%d%d", &u, &v, &w); u--, v--;
for(int j=0; j<2; j++) {
if(dis1[u] + w + dis2[v] < Time) {
Ticket = u;
Time = dis1[u] + w + dis2[v];
path = paths1[u];
for(int l=paths2[v].size() - 1; l>=0; l--)
path.push_back(paths2[v][l]);
}
swap(u, v);
}
}
if(Case++ > 0) printf("\n");
for(int i=0; i<(int)path.size(); i++)
printf("%d%c", path[i] + 1, i == (int)path.size()-1 ? '\n' : ' ');
if(Ticket == -1) printf("Ticket Not Used\n"); else printf("%d\n", Ticket + 1);
printf("%d\n", Time);
}
return 0;
}