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>
#include <limits.h>
#define M 10000
#define E 500000
   
typedef struct edge {
    int from, to, cost;
} Ed;
Ed e[E];
int d[M];
   
void sp(int,int);
  
int main(int argc,char* argv[])
{
    int vn, en, an;
    int i, j, vi;
    scanf("%d", &vn);
   
    en=0;
    for(i=0; i<vn; i++) {
        scanf("%d", &vi);
        scanf("%d", &an);
        for(j=1; j<=an; j++) {
            e[en].from=vi;
            scanf("%d %d", &e[en].to, &e[en].cost);
            en++;
        }
    }
       
    sp(vn, en);
    for(i=0; i<vn; i++) {
        printf("%d %d\n", i, d[i]);
    }
    return 0;
}
void sp(int vn, int en)
{
    int i, new=1;
    for(i=1; i<vn; i++) d[i]=INT_MAX;
       
    while(new) {
        new=0;
        for(i=0; i<en; i++) {
            if(d[e[i].from] !=INT_MAX && d[e[i].from]+e[i].cost < d[e[i].to]) {
                d[e[i].to] = d[e[i].from]+e[i].cost;
                new++;
            }
        }
    }
}