Problem:
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Single Source Shortest Path

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

$1 \leq n \leq 100$

$0 \leq c_i \leq 100,000$

$|E| \leq 10,000$

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: Time Limit Exceeded)

#include<stdio.h>
#define N 105
#define WHITE 0
#define BLACK 1
#define inf 1000000

int a[N][N];
int n;

void sssp();
int main(){

  int i,j,k,v,c;

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

  sssp();

  return 0;
}

void sssp(){
  int u,b,v;
  int cost,s=0;
  int d[N],p[N],color[N];

  for(u=0; u < n; u++){
    d[u]=inf;
    p[u]=0;
    color[u]=WHITE;
  }
  d[0]=0;

  while(1){
    cost=inf;
    for(b=0; b < n; b++){
      if(color[b] ==WHITE && d[b] < cost){
	cost=d[b];
	u=b;
      }
    }
      if(cost == inf) break;

      color[u]=BLACK;

      for(v=0; v < n; v++){
	if(color[v] == WHITE && d[u]+a[u][v] < d[v]){
	  p[v]=u;
	  d[v]=d[u]+a[u][v];
	}
      }
  }

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