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 fm;
	int to;
	int ct;
};
typedef struct edge_t Edge;

int main(void) {
	int n, i, k, fm, to, ct, upd, nedge = 0;
	int dis[10000];
	Edge *edges, *p;

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

	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d %d", &fm, &k);

		while (k--) {
			scanf("%d %d", &to, &ct);

			p = &edges[nedge++];
			p->fm = fm;
			p->to = to;
			p->ct = ct;
		}
		dis[i] = INT_MAX;
	}
	dis[0] = 0;

	upd = 1;
	while (upd) {
		upd = 0;
		for (i = 0; i < nedge; i++) {
			fm = edges[i].fm;
			to = edges[i].to;
			ct = edges[i].ct;
			if (dis[fm] < INT_MAX && dis[to] > ct + dis[fm]) {
				dis[to] = ct + dis[fm];
				upd = 1;
			}
		}
	}
	for (i = 0; i < n; i++) {
		printf("%d %d\n", i, dis[i]);
	}

	
	return 0;
}