AOJ GRL_1_A: Single Source Shortest Path （Dijktra算法求单源最短路径，邻接表）

题目链接：http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=GRL_1_A

Single Source Shortest Path

Input

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.

Output

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.

Constraints

1 ≤ |V| ≤ 100000
0 ≤ di ≤ 10000
0 ≤ |E| ≤ 500000
There are no parallel edges
There are no self-loops

Sample Input 1

4 5 0
0 1 1
0 2 4
1 2 2
2 3 1
1 3 5

Sample Output 1

0
1
3
4

Sample Input 2

4 6 1
0 1 1
0 2 4
2 0 1
1 2 2
3 1 1
3 2 5

Sample Output 2

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 2147483647struct 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;}