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>
#include<stdlib.h>

static MAX = 10001;
static const int INFTY = (1<<21);

main(){
  int n, i, j, e, sum,k,a,v1;
  int key,pi[MAX],v[MAX],u,c[MAX],**M;

  scanf("%d", &n);
 
  M=(int **)malloc(sizeof(int)*(MAX));
  for ( i = 0; i< n; i++ ){
    M[i]=(int *)malloc(sizeof(int)*(MAX));
  }
  for ( i = 0; i< n; i++ ){
    for ( j = 0; j < n; j++ ){
      M[i][j] = INFTY;
    }
  }
 
  for ( i = 0; i < n; i++ ){
    scanf("%d %d", &u, &k);
    for ( j = 0; j < k; j++ ){
      scanf("%d %d", &v1, &a);
      M[u][v1] = a;
    }
  }
  
  for(i=0;i<n;i++){
    c[i]=INFTY;
    pi[i]=-1;
    v[i]=0;
  }
  c[0]=0;

  while(1){
    key=INFTY;
    u=-1;
    for(i=0;i<n;i++){
      if(key>c[i] && !v[i]){
	u=i;
	key=c[i];
      }
    }
    if(u==-1)break;
    v[u]=1;
    for(j=0;j<n;j++){
      if(M[u][j]+c[u]<c[j] && !v[j]){
	pi[j]=u;
	c[j]=c[u]+M[u][j];
      }
    }
  }
  
  for ( i = 0; i < n; i++ ){
    printf("%d %d\n", i, (c[i]==INFTY?-1:c[i]));
  }
  
  return 0;
}