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: Wrong Answer)

#include<stdio.h>
int a[10001][10001];
int d[10001];
int pi[10001];
char color[10001];
int n,mincost;

void dijkstra(void);

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]=0;
    }
  }

  for(i=0;i<n;i++){
    scanf("%d %d" ,&u,&k);      
  for(j=1;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 i,j,u;
  for(i=0;i<n;i++){
    d[i]=100000000;
    pi[i] = -1;
    color[i]='W';
  }
  d[0]=0;
  while(1){
    mincost = 50000000;
    for(i=0;i<n;i++){
      if(color[i]!='B' && d[i] < mincost){
	mincost = d[i];
        u = i;
      }
    }
    if(mincost==50000000) break;
    color[u]='B';
  
  for(i=0;i<n;i++){
    if(color[i]!='B' && a[u][i]!=0  && d[u]+a[u][i]<d[i]){
      pi[i] = u;
      d[i]=d[u]+a[u][i];
     }
  }
  }
}