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: Memory Limit Exceeded)

#include <stdio.h>
#define INF 200000
#define WHITE 0
#define GRAY 1
#define BLACK 2
#define MAX 10000
int n;
int color[MAX];
int M[MAX][MAX];
int d[MAX];
int p[MAX];
void Dij(){
        int u,Minv;
        int i,v;
        int sum=0;
        for(i=0; i<n; i++) {
                color[i]=WHITE;
                d[i]=INF;
                p[i]=-1;

        }
        d[0]=0;
        while(1) {
                Minv=INF;
                for(i=0; i<n; i++) {
                        if(color[i]!=BLACK && d[i]<Minv) {
                                Minv = d[i];
                                u=i;
                        }
                }
                if (Minv == INF) {
                        break;
                }
                color[u]=BLACK;
                for(v=0; v<n; v++) {
                        if(color[v]!=BLACK && M[u][v]!=INF) {
                                if(d[u]+M[u][v]<d[v]) {
                                        d[v]=d[u]+M[u][v];
                                        p[v]=u;
                                        color[v]=GRAY;
                                }
                        }
                }

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

}
int main(){
        int i,j;
        int k,c,u,v;
        scanf ("%d",&n);
        for(i=0; i<n; i++) {
                for(j=0; j<n; j++) {
                        M[i][j]=INF;
                }
        }

        for(i=0; i<n; i++) {
                scanf("%d%d",&u,&k);
                for(j=0; j<k; j++) {
                        scanf("%d%d",&v,&c);
                        M[u][v]=c;
                }
        }
        Dij();
        return 0;
}