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 JavaScript (judge verdict: Memory Limit Exceeded)

var input = require('fs').readFileSync('/dev/stdin', 'utf8').trim();
var lines = input.split('\n');

var n = +lines.shift();
var G = [];
for (var i = 0; i < n; i++) {
    G.push([]);
    for (var j = 0; j < n; j++) {
        G[i][j] = Number.POSITIVE_INFINITY;
    }
    G[i][i] = 0;
}
for (var i = 0; i < n; i++) {
    var u = lines.shift().trim().split(' ').map(function(n){return +n});
    var id = u[0];
    var k = u[1];
    for (var j = 0; j < k; j++) {
        var adj = u[2 + j * 2];
        var cost = u[2 + j * 2 + 1];
        G[i][adj] = cost;
    }
}

var cost = [0];
for (var i = 1; i < n; i++) {
    cost.push(Number.POSITIVE_INFINITY);
}
var visited = [];
while (true) {
    var u = -1;
    for (var i = 0; i < n; i++) {
        if (!visited[i] && (u == -1 || cost[i] < cost[u])) {
            u = i;
        }
    }
    if (u == -1) {
        break;
    }
    visited[u] = true;

    for (var i = 0; i < n; i++) {
        cost[i] = Math.min(cost[i], cost[u] + G[u][i]);
    }
}

for (var i = 0; i < n; i++) {
    console.log(i, cost[i]);
}