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 g;
  int b;
  int c;
}data;

void single(int);

int n,A[N];
data G[500000];

int main()
{
  int i,u,k,j,cnt = 0;

  scanf("%d",&n);

  for(i = 0; i < n; i++){
    A[i] = INF;
    A[0] = 0;
  }
    
  for(i = 0; i < n; i++){
    scanf("%d %d",&u,&k);
    for(j = 0; j < k; j++){
      G[cnt].b = u;
      scanf("%d%d",&G[cnt].g, &G[cnt].c);
      cnt++;
    }
  }
  single(cnt);

  for(i = 0; i < n; i++){
    printf("%d %d\n",i,A[i]);
  }

  return 0;
}

void single(int num){
  int i,flag = 1;

  for(i = 0; i < n; i++){
    A[i] = INF;
    A[0] = 0;
  }

    while(flag != 0){
      flag = 0;
      for(i = 0; i < num; i++){
	if(G[i].c + A[G[i].b] < A[G[i].g]){
	  A[G[i].g] = G[i].c + A[G[i].b];
	  flag = 1;
	}
      }
    }
}