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>

typedef struct edge_t{
	int from;
	int to;
	int heavy;
}edge;

int main(void) {
	int n,i,k,from,to,heavy,hantei,nedge=0,dis[10000];
	edge *edges, *p;

	edges = malloc(sizeof(edge)*500000);

	scanf("%d", &n);

	for (i = 0; i < n; i++) {
		scanf("%d",&from);
    scanf("%d",&k);
		while (k--) {
			scanf("%d",&to);
			scanf("%d",&heavy);
			p = &edges[nedge++];
			p->from = from;
			p->to = to;
			p->heavy = heavy;
		}
		dis[i] = INT_MAX;
	}
	dis[0] = 0;
	hantei = 1;
	while (hantei) {
		hantei = 0;
		for (i = 0; i < nedge; i++) {
			from = edges[i].from;
			to = edges[i].to;
		  heavy = edges[i].heavy;
			if (dis[from] < INT_MAX && dis[to] > heavy + dis[from]) {
				dis[to] = heavy + dis[from];
				hantei = 1;
			}
		}
	}

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