Basic method for basic storage of graphs three

Basic method for basic storage of graphs three Time limit:1000ms Memory limit:65536k in doubt. Dot here ^_^ Title Description

To solve the problem of graph theory, we must first think about how to store the graph. But Xiao Xin but how also can not understand how to save a map to help solve the problem. Can you help him to solve the problem? Enter multiple sets of inputs to the end of the file.

The first row of each group has two numbers n, m represents n points, and m bars have a forward edge. Next there are m lines, two numbers per line, U, V, W for u to V with a forward Benquan value of W. Line M+2 has a number Q for the number of queries, followed by Q line each line has a query, enter a number of a

Note: The point number is 0~n-1,2<=n<=500000, and 0<=m<=500000,0<=q<=500000,u!=v,w is the int type data. Input guarantees no self-loop and heavy-side output for each query, the output line is two x, Y. Indicates that the first edge of a sort is x through Y. For each edge, the collation is as follows:

The weights are small in front.

The weight of the equal edge starting point number small in the preceding weight and starting point equal arrival point number small in the front

Note: The edges are numbered from 0 to start the sample input

4 3
0 1 1
1 2 2
1 3 0
3
0
1
2

Sample Output

1 3
0 1
1 2

Tips SourceLin Sample Program

#include <stdio.h>  
#include <stdlib.h>  
#define MAX 2147483647  
struct node  
{  
int u;  
int V;  
int w;  
} S[500010];  
int cmp (const void *a,const void *b)  
{  
struct node *c= (struct node *) A;  
struct node *d= (struct node *) b;  
if (c->w!=d->w)  
return c->w-d->w;  
else if (c->u!=d->u)  
return c->u-d->u;  
else  
return c->v-d->v;  
}  
int main ()  
{  
int i,j,n,m,k,t,l1,l2,l3;  
while (scanf ("%d%d", &n,&m)!=eof)  
{for  
(i=0;i<n;i++)  
S[i].w=max;  
for (i=0;i<m;i++)  
{  
scanf ("%d%d%d", &l1,&l2,&l3);  
S[I].U=L1;  
S[I].V=L2;  
s[i].w=l3;  
}  
Qsort (S,m,sizeof (s[0]), CMP);  
scanf ("%d", &k);  
while (k--)  
{  
scanf ("%d", &t);  
printf ("%d%d\n", S[T].U,S[T].V);}}}