hdu 4707 Pet（DFS && 邻接表）

Pet

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1809 Accepted Submission(s): 874



[align=left]Problem Description[/align]
One day, Lin Ji wake up in the morning and found that his pethamster escaped. He searched in the room but didn’t find the hamster. He tried to use some cheese to trap the hamster. He put the cheese trap in his room and waited for
three days. Nothing but cockroaches was caught. He got the map of the school and foundthat there is no cyclic path and every location in the school can be reached from his room. The trap’s manual mention that the pet will always come back if it still in somewhere
nearer than distance D. Your task is to help Lin Ji to find out how many possible locations the hamster may found given the map of the school. Assume that the hamster is still hiding in somewhere in the school and distance between each adjacent locations is
always one distance unit.

[align=left]Input[/align]
The input contains multiple test cases. Thefirst line is a positive integer T (0<T<=10), the number of test cases. For each test cases, the first line has two positive integer N (0<N<=100000) and D(0<D<N), separated by a single space.
N is the number of locations in the school and D is the affective distance of the trap. The following N-1lines descripts the map, each has two integer x and y(0<=x,y<N), separated by a single space, meaning that x and y is adjacent in the map. Lin Ji’s room
is always at location 0.

[align=left]Output[/align]
For each test case, outputin a single line the number of possible locations in the school the hamster may be found.

[align=left]Sample Input[/align]

1
10 2
0 1
0 2
0 3
1 4
1 5
2 6
3 7
4 8
6 9

[align=left]Sample Output[/align]

2

[align=left]Source[/align]
2013 ACM/ICPC Asia Regional Online —— Warmup

求全部房子建一颗树，到根节点距离大于 D的节点的个数

DFS。邻接表存储，，。第一次用邻接表。对于菜鸟的我，

第一次接触还真不好理解。我把我理解半天的邻接表存储的都凝视了

希望对小伙伴们理解有帮助，，

#include<stdio.h>
#include<string.h>
#define M 100100
struct node{
int to,next,w;//w为边权值
}edge[M];
int head[M];
int cnt,n,m,ans;
void add(int x,int y){//邻接表是对每一条边进行编号，从0開始。
edge[cnt].to=y;//edg[cnt].to表示编号为cnt的那条边的终点
edge[cnt].next=head[x];//edge[cnt].next表示和cnt这条边同起点的下一条边的编号
head[x]=cnt++;//head[i]的意思是以i为起点的第一条边的编号，（ 事实上就是以i为起点最后输入的那条边的编号，就是逆序第一条边的编号）
}
void dfs(int x,int w){
if(head[x]==-1)
return;
for(int i=head[x]; i!=-1 ;i=edge[i].next){//从以x为起点的最后一个输入的边開始遍历
//i=edge[i].next表示和上一条边同起点的下一条边的编号，依次遍历
int to=edge[i].to;
edge[i].w=w+1;
if(edge[i].w > m)
ans++;
dfs(to,edge[i].w);
}
}
int main(){
int t,i,a,b;
scanf("%d",&t);
while(t--){
memset(head,-1,sizeof(head));
scanf("%d%d",&n,&m);
cnt=0;
for(i=0;i<n-1;i++){
scanf("%d%d",&a,&b);
add(a,b);
}
ans=0;
dfs(0,0);
printf("%d\n",ans);
}
return 0;
}