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>
#define INF 100000000
#define NIL -1
#define WHITE 0
#define BLACK 1

typedef struct nodes{
	int number;
	int d;
	struct nodes *next;
}Node;

void addNode(Node*, int, int);

main(){
	int n, i, j, k, m, l, u, *d, *flag, mincost;
	Node *G, *p;

	scanf("%d", &n);
	d = (int *)malloc(sizeof(int) * n);
	flag = (int *)malloc(sizeof(int) * n);
	G = (Node *)malloc(sizeof(Node) * n);
	for (i = 0; i < n; i++){
		G[i].next = &(G[i]);  //??£??\?????°0
		d[i] = INF;
		flag[i] = WHITE;
		scanf("%*d%d", &k);
		for (j = 0; j < k; j++){
			scanf("%d%d", &m, &l);
			addNode(&(G[i]), m, l);
		}
	}

#ifdef DEBUG
	printf("*********************\n");
	for (i = 0; i < n; i++){
		printf("[%d] ", i);
		for (p = G[i].next; p != &(G[i]); p = p->next){
			printf("%d-%d , ", p->number, p->d);
		}
		printf("\n");
	}
	printf("*********************\n");
#endif

	d[0] = 0;  // ?§?????????????

	while (1){
		mincost = INF;
		for (i = 0; i < n; i++){
			if (flag[i] != BLACK && d[i] < mincost){
				mincost = d[i];
				u = i;
			}
		}
		if (mincost == INF)break; // ??¢?´¢??????(??¨???BLACK)
		flag[u] = BLACK;

		for (p = G[u].next; p != &(G[u]); p = p->next){
			if (d[p->number] > d[u] + p->d){
				d[p->number] = d[u] + p->d;
			}
		}
	}

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

	return 0;
}

void addNode(Node*graph, int k, int d){
	Node* p;

	for (p = graph->next; p->next != graph; p = p->next);
	p->next = malloc(sizeof(Node));
	p->next->next = graph;
	p->next->number = k;
	p->next->d = d;

#ifdef DEBUG
	printf("**add** %d %d \n", p->number, p->d);
#endif
}