标签：表示 online ack font targe ffffff ref input constrain
题目链接：http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=GRL_1_A
An edge-weighted graph G (V, E) and the source r.
|V| |E| r s0 t0 d0 s1 t1 d1 : s|E|?1 t|E|?1 d|E|?1
|V| is the number of vertices and |E| is the number of edges in G. The graph vertices are named with the numbers 0, 1,..., |V|?1 respectively. r is the source of the graph.
si and ti represent source and target vertices of i-th edge (directed) and di represents the cost of the i-th edge.
c0 c1 : c|V|?1
The output consists of |V| lines. Print the cost of the shortest path from the source r to each vertex 0, 1, ... |V|?1 in order. If there is no path from the source to a vertex, print INF.
4 5 0 0 1 1 0 2 4 1 2 2 2 3 1 1 3 5
0 1 3 4
4 6 1 0 1 1 0 2 4 2 0 1 1 2 2 3 1 1 3 2 5
3 0 2 INF
这题数据范围比较大，最多有100000个顶点，用邻接矩阵表示的话，空间肯定会超限。
我用Dijiksra的邻接表来解了。
套一个模板就出来了：
#include <iostream> #include <algorithm> #include <map> #include <vector> using namespace std; typedef long long ll; #define INF 2147483647 struct edge{ int to,cost; }; int V,E; vector <edge> G[100010]; multimap <int,int> l; ll d[500010]; void dijkstra(int s){ fill(d,d+V,INF); d[s] = 0; l.insert(make_pair(0,s)); while(l.size() > 0){ int p = l.begin()->first; int v = l.begin()->second; l.erase(l.begin()); if(d[v] < p) continue; for(int i = 0;i < G[v].size(); i++){ edge e = G[v][i]; if(d[e.to] > d[v] + e.cost){ d[e.to] = d[v] + e.cost; l.insert(make_pair(d[e.to], e.to)); } } } } int main(){ int r; cin >> V >> E >> r; for(int i = 0;i < E; i++){ edge e; int from; cin >> from >> e.to >> e.cost; G[from].push_back(e); } dijkstra(r); for(int i = 0;i < V; i++){ if(d[i] != INF) cout << d[i] << endl; else cout << "INF" <<endl; } return 0; }
AOJ GRL_1_A: Single Source Shortest Path （Dijktra算法求单源最短路径，邻接表）
标签：表示 online ack font targe ffffff ref input constrain
原文地址：http://www.cnblogs.com/zhangjiuding/p/7726089.html
