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>
 
struct edge_t {
    int from;
    int to;
    int cost;
};
typedef struct edge_t Edge;
 
int main(void) {
    int num, i, k, from, to, cost, update, nedge = 0;
    int dis[10000];
    Edge *edges, *p;
 
    edges = malloc(sizeof(Edge)*500000);
 
    scanf("%d", &num);
 
    for (i = 0; i < num; i++) {
        scanf("%d %d", &from, &k);
 
        while (k--) {
            scanf("%d %d", &to, &cost);
 
            p = &edges[nedge++];
            p->from = from;
            p->to = to;
            p->cost = cost;
        }
 
        dis[i] = INT_MAX;
    }
    dis[0] = 0;
 
    update = 1;
    while (update) {
        update = 0;
        for (i = 0; i < nedge; i++) {
            from = edges[i].from;
            to = edges[i].to;
            cost = edges[i].cost;
            if (dis[from] < INT_MAX && dis[to] > cost + dis[from]) {
                dis[to] = cost + dis[from];
                update = 1;
            }
        }
    }
 
    for (i = 0; i < num; i++) {
        printf("%d %d\n", i, dis[i]);
    }
    return 0;
}