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 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;
}