UVA - 10986 Sending email （Dijkstra邻接表+优先队列优化）

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题目大意:
给一个图，求从s点到t点的最小距离。

解析：
赤裸裸的最短路，但n太大显然是不能用邻接矩阵的，需要对Dijkstra用邻接表+优先队列优化。

这里我把Dijkstra的邻接表+优先队列的方法封装成了一个类，感觉挺好用的。

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#include <queue>using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 20005;struct Edge {int from, to, dist;Edge(int u,int v,int d) {from = u;to = v;dist = d;}};struct HeapNode {int d, u;HeapNode(int _d,int _u) {d = _d;u = _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]; //是否被永久标记int d[maxn]; //s到各个点的距离int p[maxn]; //最短路上的一条弧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] = G[u][i];Q.push(HeapNode(d[e.to], e.to));}}}}};int main() {Dijkstra dij;int t,cas = 1;int n,m;int u,v,dist;int s,e;scanf("%d",&t);while(t--) {scanf("%d%d%d%d",&n,&m,&s,&e);dij.init(n);for(int i = 0; i < m; i++) {scanf("%d%d%d",&u,&v,&dist);dij.AddEdge(u,v,dist);dij.AddEdge(v,u,dist);}dij.dijkstra(s);printf("Case #%d: ",cas++);if(dij.d[e] == INF) {printf("unreachable\n");}else {printf("%d\n",dij.d[e]);}}return 0;}

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