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>
#define MAX 10000
#define INF 1000000
#define NIL -1
 
int a[MAX][MAX];
int n;
int d[MAX];
int pi[MAX];
 
void R(int, int);
void ISS(int);
void D(int);
 
main(){
 
  int i, j ,u ,k,v,c;
  scanf("%d",&n);
 
  for(i=0; i<n; i++){
    for(j=0; j<n; j++){
      a[i][j]=-1;
    }
  }
 
  for(i=0; i<n; i++){
    scanf("%d %d",&u,&k);
    for(j=0; j<k; j++){
      scanf("%d %d",&v,&c);
      a[i][v]=c;
    }
  }
 
  D(0);
  for(i=0; i<n; i++){
    printf("%d %d\n", i, d[i]);
  }
 
  return 0;
}
 
void R(int u, int v){
  if(d[v]> d[u]+a[u][v]){
    d[v]=d[u]+a[u][v];
    pi[v]=u;
  }
}
 
void ISS(int s){
  int v;
  for(v=0; v<n; v++){
    d[v]=INF;
    pi[v]=NIL;
  }
  d[s]=0;
}
 
void D(int s){
  int i, v1, v, minv, u;
  int flag[MAX];
  ISS(s);
  
  for(i=0; i<n; i++) flag[i]=0;
 
  for(v1=0; v1<n; v1++){
    
    minv = INF;
    for ( i = 0; i < n; i++ ){
      if(flag[i]==1) continue;
      if(d[i]<=minv){
    u=i;
    minv=d[i];
      }
    }
    
    for(v=0; v<n; v++){
      if(a[u][v]==-1) continue;
      if(flag[v]==1) continue;
 
      R(u, v);
    }
    flag[u]=1;
  }
}