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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#define MAX 10000
struct node{
  int key;
  int cost;
  struct node *next;
};

typedef struct node *adj;
adj G[MAX];
adj SuperNode(int key,int cost,adj Node){
  adj n;
  n=(adj)malloc(sizeof(struct node));
  n->key=key;
  n->cost=cost;
  n->next=Node;
  return n;
}

adj Insert(int a,int b,int c){
  adj H;
  H=SuperNode(b,c,G[a]->next);
  G[a]->next=H;
}

int main(){
  int i,j,n,count=0,v,c,c1,v1;
  int color[MAX],D[MAX],pi[MAX],nx,min;
  adj H;

  scanf("%d",&n);
  for(i = 0;i < n;i++)G[i]=SuperNode(0,0,NULL);
  for(i = 0;i < n;i++){
    scanf("%d%d",&v,&c);

    for(j = 0;j <c;j++){
      scanf("%d %d",&v1,&c1);
    Insert(i,v1,c1);
  }
}

for(i=0;i<n;i++){
  color[i]=0;
  pi[i]=-1;
  D[i]=10000000;
 }
 D[0]=0;

while(1){
  min=10000000;
  for(i=0;i<n;i++){
    if(color[i] != 2 && D[i] < min){
      min=D[i];
      nx=i;
    }
  }

  if(min==10000000)break;
  color[nx]=2;

  for(H=G[nx];H!=NULL;H=H->next){
    if(color[H->key] !=2 && (H->cost+D[nx]) < D[H->key]){
        pi[H->key]=nx;
        D[H->key]=H->cost+D[nx];
        color[H->key]=1;
    }

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