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 INF 10000000
#define NIL -1
#define BLACK 1
#define WHITE 0

void dijkstra(void);

int n,a[10001][10001],w=0,d[10001],color[10001];

int main()
{
  int i,j,u,k,v;


  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",&u); //頂点の番号
    scanf("%d",&k); //出次数

    for(j=0;j<k;j++){
      scanf("%d",&v);
      scanf("%d",&a[u][v]);
    }
  }

  dijkstra();

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

  return 0;
}

void dijkstra(void){
  int mincost,u,v,i;

  for(u=0;u<n;u++){ // each u in G
    d[u] = INF;
    color[u] = WHITE;
  }

  d[0] = 0;

  while(1){
    mincost = INF;
    for(i=0;i<n;i++){ //each i in G
      if (color[i] != BLACK && d[i] < mincost){
	mincost = d[i];
	u = i;
      }
    }
    color[u] = BLACK;

    if(mincost == INF) break;

    for(v=0;v<n;v++){// each v in Adj[u]
      if(a[u][v]!=-1){
	if (color[v] != BLACK && a[u][v]+d[u] < d[v]){
	  d[v] = d[u]+a[u][v];
	}
      }
    }
  }
}