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: Memory Limit Exceeded)

#include <stdio.h>

int a,b, c,n,u,w[10000][10000],color[10000],d[10000];

int main(){
  int i,j,mincost,sum=0;
  scanf("%d",&n);
 for(i=0;i<n;i++)
   for(j=0;j<n;j++)w[i][j]=100008;

  for(i=0;i<n;i++){
    scanf("%d%d",&a,&c);
      for(j=0;j<c;j++){
scanf("%d",&b);
 scanf("%d",&w[a][b]);    
}
    }
 
  for(i=0;i<n;i++){
    d[i] =100008;
    color[i] =0;
  }
  d[0]=0;

  while (1){
    mincost=100008;    
for(i=0;i<n;i++){
      if (color[i] != 1 && d[i] < mincost){
	mincost = d[i];
	u = i;
      }
    }
    if (mincost ==100008)
      break;
    color[u] = 1;
    for (i=0;i<n;i++){
      if (color[i] != 1 && d[u]+w[u][i] < d[i]){
	d[i] = d[u]+w[u][i];
      }
    }
  }
 
 for (i=0;i<n;i++)
 printf("%d %d\n",i,d[i]); 

  return 0;
}