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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

struct vertex {
  int k;
  int dis;
  int definite;
  int **vlist;
};

int main(int argc, char *argv[])
{
  int n, i, j, id;
  struct vertex *adj;
  int minv, op, cost, mind;

  scanf("%d", &n);
  adj = (struct vertex *)malloc(sizeof(struct vertex) * n);

  for (i = 0; i < n; i++) {
    scanf("%d", &id);
    scanf("%d", &(adj[id].k));
    adj[id].vlist = (int **)malloc(sizeof(int *) * adj[id].k);
    for (j = 0; j < adj[id].k; j++) {
      adj[id].vlist[j] = (int *)malloc(sizeof(int) * 2);
      scanf("%d", &(adj[id].vlist[j][0]));
      scanf("%d", &(adj[id].vlist[j][1]));
    }
    adj[id].dis = INT_MAX;
    adj[id].definite = 0;
  }

  adj[0].dis = 0;
  adj[0].definite = 1;
  minv = 0;
  while (minv != -1) {
    for (i = 0; i < n; i++) {
      if (adj[i].definite == 1) {
	for (j = 0; j < adj[i].k; j++) {
	  op = adj[i].vlist[j][0];
	  cost = adj[i].vlist[j][1];
	  if (adj[op].definite == 0 && adj[op].dis > adj[i].dis + cost) {
	    adj[op].dis = adj[i].dis + cost;
	  }
	}
      }
    }
    minv = -1;
    mind = INT_MAX;
    for (i = 0; i < n; i++) {
      if (adj[i].definite == 0 && mind > adj[i].dis) {
	mind = adj[i].dis;
	minv = i;
      }
    }
    if (minv != -1) adj[minv].definite = 1;
  }


  for (i = 0; i < n; i++) printf("%d %d\n", i, adj[i].dis);

  return 0;
}