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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

#define INF (INT_MAX/2)


typedef struct {
  short v; // 接続先の番号
  int c; // 有向辺の重み
} edge;

typedef struct {
  int k; // 頂点の持つ有向辺の数
  edge* e; // 有向辺の配列
} vertex;



int* ssspForCost( vertex* g, int n ) {

  int i;
  int *dist, color[n];
  int minCost, u;

  dist = (int*)malloc( sizeof( int ) * n );

  //初期化
  for( i=0; i<n; i++ ) { 
    dist[i] = INF;
    color[i] = 0;
  }

  dist[0] = 0;

  while( 1 ) {

    minCost = INF;
    for( i=0; i<n; i++ ) {
      if( color[i] != 1 && dist[i] < minCost ) {
	minCost = dist[i];
	u = i;
      }
    }

    if( minCost == INF ) {
      break;
    }

    for( i=0; i<g[u].k; i++ ) {
      if( color[i] != 1 && dist[u] + g[u].e[i].c < dist[i] ) {
	dist[i] = dist[u] + g[u].e[i].c;
      }
    }

  }

  return dist;

}


void printDist( int* dist, int n ) {
  int i;

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

  return;

}


int main() {

  int i, j;
  int n, u; // n:頂点の数
  vertex* g;
  
  int* dist;

  scanf( "%d", &n );
  g = (vertex*)malloc( sizeof( vertex ) * n );

  while( 1 ) {
    if( scanf( "%d", &u ) == EOF ) {
      break;
    }
    scanf( "%d", &g[u].k );
    g[u].e = (edge*)malloc( sizeof( edge ) * g[u].k );

    for( j=0; j<g[u].k; j++ ) {
      scanf( "%hd%d", &g[u].e[j].v, &g[u].e[j].c );
    }
  }

  dist = ssspForCost( g, n );

  printDist( dist, n );

  return 0;
}