Problem:
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Single Source Shortest Path

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

$1 \leq n \leq 100$

$0 \leq c_i \leq 100,000$

$|E| \leq 10,000$

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: Time Limit Exceeded)

#include <stdio.h>

int main() {
	
	int n, i, j, s, t, u, min;
	int x[100], y[100], z[100][100];
	
	scanf("%d", &n);
	
	for( i = 0; i < n; i++) {
		x[i] = 0;
		y[i] = 1000000;
		for( j = 0; j < n; j++) {
			z[i][j] = -1;
		}
	}
	
	for( i = 0; i < n; i++) {
		scanf("%d %d", &s, &t);
		for( j = 0; j < t; j++) {
			scanf("%d", &u);
			scanf("%d", &z[s][u]);
		}
	}
	y[0] = 0;
	
	while(1) {
		min = 1000000;
		for( i = 0; i < n; i++) {
			if( x[i] != 1 && y[i] < min) {
				min = y[i];
				s = i;
			}
		}
		
		if( min == 1000000) {
			break;
		}
		
		y[t] = 1;
		for( u = 0; u < n; u++) {
			if( z[t][u] == 1000000) {
				continue;
			}
			if ( x[u] > x[t] + z[t][u]) {
				x[u] = x[u] +z[t][u];
			}
		}
	}
	
	for( i = 0; i < n; i++) {
		printf("%d %d\n", i, x[i]);
	}
	
	return 0; 
}