PTA 7-10 旅游规划

PTA 7-10 旅游规划

有了一张自驾旅游路线图，你会知道城市间的高速公路长度、以及该公路要收取的过路费。现在需要你写一个程序，帮助前来咨询的游客找一条出发地和目的地之间的最短路径。如果有若干条路径都是最短的，那么需要输出最便宜的一条路径。

#include #include #include #include #include using namespace std;typedef pair P;const int INF = 0x3f3f3f3f;const int maxn = 550;const int maxm = maxn*maxn;int tot = 0, head[maxn];void init() { tot = 0; memset(head, -1, sizeof(head));}struct Edge { int from, to, dis, cost, next; Edge(int f = 0, int t = 0, int d = 0, int c = 0, int n = -1) : from(f), to(t), dis(d), cost(c), next(n) {}}edge[maxm];void addedge(int u, int v, int dis, int cost) { ++tot; edge[tot].to = v, edge[tot].dis = dis, edge[tot].cost = cost; edge[tot].next = head[u]; head[u] = tot;}int n, m, s, e;bool vis[maxn];int sdis[maxn], scost[maxn];void dijkstra() { for (int i = 0; i <= n; i++) sdis[i] = INF, scost[i] = 0, vis[i] = false; sdis[s] = 0; priority_queue, greater

> q; q.push(make_pair(0, s)); while (!q.empty()) { int u = q-().second; q.pop(); if (vis[u]) continue; vis[u] = true; for (int i = head[u]; i != -1; i = edge[i].next) { int v = edge[i].to, dis = edge[i].dis, cost = edge[i].cost; if (sdis[v] > sdis[u] + dis) { sdis[v] = sdis[u] + dis; scost[v] = scost[u] + cost; q.push(make_pair(sdis[v], v)); } else if (sdis[v] == sdis[u] + dis && scost[v] > scost[u] + cost) { scost[v] = scost[u] + cost; } } }}int main() { scanf("%d %d %d %d", &n, &m, &s, &e); init(); int u, v, cost, dis; for (int i = 1; i <= m; i++) { scanf("%d %d %d %d", &u, &v, &dis, &cost); addedge(u, v, dis, cost); addedge(v, u, dis, cost); } dijkstra(); printf("%d %d\n", sdis[e], scost[e]);}

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