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



#define N 100000
#define INF 1000000

typedef struct{
  int go;
  int back;
  int cost;
}Date;

int n, d[N];
Date a[500000];
void root(int);

int main(){
  int i, j, l, k, count = 0;
  scanf("%d", &n);
  for( i = 0 ; i < n ; i++) {
    scanf("%d%d", &l, &k);
    for( j = 0 ; j < k ; j++){
      a[count].back=l;
      scanf("%d%d", &a[count].go, &a[count].cost);
      count++;
    }
  }






  root(count);
  for( i = 0 ; i < n ; i++) printf("%d %d\n", i, d[i]);
  return 0;
}






void root(int b){
  int j, f = 1;

  for( j = 0 ; j < n ; j++) d[j] = INF;
  d[0] = 0;

  while(f != 0){
    f = 0;
    for( j = 0 ; j < b ; j++){
      if(a[j].cost + d[a[j].back] < d[a[j].go]){
        d[a[j].go] = a[j].cost + d[a[j].back];
        f = 1;
      }
    }
  }
}