Problem:
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Single Source Shortest Path II

For a given weighted graph $G = (V, E)$, find the shortest path from a source to each vertex. For each vertex $u$, print the total weight of edges on the shortest path from vertex $0$ to $u$.

Input

In the first line, an integer $n$ denoting the number of vertices in $G$ is given. In the following $n$ lines, adjacency lists for each vertex $u$ are respectively given in the following format:

$u$ $k$ $v_1$ $c_1$ $v_2$ $c_2$ ... $v_k$ $c_k$

Vertices in $G$ are named with IDs $0, 1, ..., n-1$. $u$ is ID of the target vertex and $k$ denotes its degree. $v_i (i = 1, 2, ... k)$ denote IDs of vertices adjacent to $u$ and $c_i$ denotes the weight of a directed edge connecting $u$ and $v_i$ (from $u$ to $v_i$).

Output

For each vertex, print its ID and the distance separated by a space character in a line respectively. Print in order of vertex IDs.

Constraints

$1 \leq n \leq 10,000$

$0 \leq c_i \leq 100,000$

$|E| < 500,000$

All vertices are reachable from vertex $0$

Sample Input 1

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

Sample Output 1

0 0
1 2
2 2
3 1
4 3

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C++ (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
struct Point{
	int v,e;
	bool operator<(const Point b)const{
		return this->e > b.e;
	}
	Point(int a,int b):v(a),e(b){}
	Point():v(0),e(0){}
};

long long d[10000];
bool visit[10000] = {0};
vector <Point> G[10000];
priority_queue <Point> PQ;

const long long INF = 1e19;

void dj(int n){
	for(int i=0; i<n; i++) d[i] = INF;
	d[0] = 0;
	PQ.push(Point(0,0));
	Point p, q;
	while(!PQ.empty()){
		p = PQ.top(); PQ.pop();
		visit[p.v] = 1;
		if(d[p.v] < p.e) continue;
		int du = G[p.v].size();
		for(int i=0; i<du; i++){
			q = G[p.v][i];
			if(!visit[q.v] && q.e+d[p.v]<d[q.v]){
				d[q.v] = q.e + d[p.v];
				PQ.push(Point(q.v,d[q.v]));
			}
		}
	}
}
int main()
{	
	int n,u,du; cin>>n;
	for(int i=0; i<n; i++){
		scanf("%d%d",&u,&du);
		int v,e;
		while(du--){
			scanf("%d%d",&v,&e);
			G[u].push_back(Point(v,e));
		}
	}
	dj(n);
	for(int i=0;i<n;i++)
		printf("%d %d\n",i,d[i]);
	return 0;
}