Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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>
typedef struct{
  int prev,next,weight;
}Data;
Data data[500000];

void SSS();
int num,size;       
int main(){
  int i,j,id,nextnum;
 
  scanf("%d",&num);
  for(i=0;i<num;i++){
    scanf("%d%d",&id,&nextnum);
    for(j=0;j<nextnum;j++){
      
      
      data[size].prev=id;
      scanf("%d%d",&data[size].next,&data[size].weight);
      size++;
     
    }
  }
  SSS();
  return 0;
}


void SSS(){
  int cost[10000];
  int i,flag=1;
  for(i=0;i<num;i++){
    cost[i]=1000001;
  }
  cost[0]=0;
 
  
  while(flag==1){
    flag=0;
    for(i=0;i<size;i++){
      if(cost[data[i].prev]==1000001) continue;
      if(cost[data[i].prev]+data[i].weight<cost[data[i].next]){
	cost[data[i].next]=cost[data[i].prev]+data[i].weight;
	flag=1;
      }
    
    
    }
  }
  
  for(i=0;i<num;i++){
    if(cost[i]==1000001) printf("%d -1\n",i);   /*output*/
    else printf("%d %d\n",i,cost[i]);
  }
}