Pet
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1754 Accepted Submission(s): 847
Problem Description
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.
Input
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.
Output
For each test case, outputin a single line the number of possible locations in the school the hamster may be found.
Sample Input
1 10 2 0 1 0 2 0 3 1 4 1 5 2 6 3 7 4 8 6 9
Sample Output
2
题意：读了半天没理解，其实就是一个图求深度大于D的数的个数；还可以用邻接表和并差集做；//前提不能成环；；；
代码：
1 #include<stdio.h> 2 #include<string.h> 3 #include<vector> 4 using namespace std; 5 vector<int>vec[100010]; 6 int dis[100010]; 7 void dfs(int x){ 8 for(int i=0;i<vec[x].size();i++){ 9 dis[vec[x][i]]=dis[x]+1;//深度加一；vec【x】【i】代表b的值也就是dis【b】等于a的深度加一； 10 dfs(vec[x][i]);//如果vec[x].size（）不为0；下一步搜索，改变深度的值；下次就是b了； 11 } 12 return; 13 } 14 int main(){ 15 int T,N,D,a,b,flot; 16 scanf("%d",&T); 17 while(T--){memset(dis,0,sizeof(dis));flot=0; 18 scanf("%d%d",&N,&D); 19 for(int i=0;i<N;i++)vec[i].clear(); 20 for(int i=0;i<N-1;i++){ 21 scanf("%d%d",&a,&b); 22 vec[a].push_back(b); 23 } 24 dfs(0); 25 for(int i=0;i<N;i++){ 26 if(dis[i]>D)flot++; 27 } 28 printf("%d ",flot); 29 } 30 return 0; 31 }
自己写的邻接表：
1 #include<stdio.h> 2 #include<string.h> 3 struct Node{ 4 int now; 5 int next; 6 int step; 7 }; 8 Node map[100010]; 9 int head[100010]; 10 int len,anser,N,D;//每次加一代表的是当前树的总结点数 11 void add(int x,int y){ 12 map[len].now=y;//now放第二个 y 13 map[len].next=head[x];//next指向第一个x；也就是y指向x; 14 head[x]=len++; 15 } 16 void dfs(int x,int st){ 17 if(st>D)anser++;//深度大于D答案就++； 18 for(int i=head[x];i!=-1;i=map[i].next){//相当于vector的size，这个以链表的查找方式； 19 map[i].step=st+1; 20 dfs(map[i].now,map[i].step);//将当前搜索到的与head[i]相接的now传下去，继续搜索； 21 } 22 return; 23 } 24 int main(){ 25 int T,a,b; 26 scanf("%d",&T); 27 while(T--){len=0;anser=0; 28 memset(map,0,sizeof(map)); 29 memset(head,-1,sizeof(head));//将head数组的初始化为-1 30 scanf("%d%d",&N,&D); 31 for(int i=0;i<N-1;i++){ 32 scanf("%d%d",&a,&b); 33 add(a,b); 34 } 35 dfs(0,0); 36 printf("%d ",anser); 37 } 38 return 0; 39 }
自己写的并查集：
1 #include<stdio.h> 2 const int MAXN=100010; 3 int tree[MAXN]; 4 int N,D; 5 int find(int x){ 6 int r=x; 7 while(r!=tree[r])r=tree[r]; 8 return r; 9 } 10 int depth(int x){int tot=0; 11 while(x!=tree[x]){ 12 x=tree[x];tot++; 13 } 14 return tot; 15 } 16 void initial(){ 17 for(int i=0;i<=N;i++)tree[i]=i; 18 } 19 void merge(int x,int y){ 20 tree[y]=x;//由于是找深度，没有路径压缩，所以直接并上； 21 } 22 int findanswer(){int answer=0; 23 for(int i=N;i>0;i--){ 24 if(depth(i)>D&&find(i)==0)answer++; 25 // printf("%d ",depth(i)); 26 } 27 return answer; 28 } 29 int main(){ 30 int T,a,b,answer; 31 scanf("%d",&T); 32 while(T--){ 33 scanf("%d%d",&N,&D); 34 initial(); 35 for(int i=0;i<N-1;i++){ 36 scanf("%d%d",&a,&b); 37 merge(a,b); 38 } 39 answer=findanswer(); 40 printf("%d ",answer); 41 } 42 return 0; 43 }
大神们用的邻接表和并差集先贴着吧；有空自己写写:
邻接表：
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 struct node 6 { 7 int next,to; 8 int step; 9 }a[100005]; 10 int head[100005]; 11 int n,d,len,ans; 12 void add(int x,int y) 13 { 14 a[len].to = y; 15 a[len].next = head[x]; 16 head[x] = len++; 17 } 18 void dfs(int x,int step) 19 { 20 int i,j,k; 21 if(-1 == head[x]) 22 return ; 23 for(i = head[x]; i!=-1; i = a[i].next) 24 { 25 k = a[i].to; 26 a[i].step = step+1; 27 if(a[i].step>d) 28 ans++; 29 dfs(k,a[i].step); 30 } 31 } 32 int main() 33 { 34 int T,i,j,x,y; 35 scanf("%d",&T); 36 while(T--) 37 { 38 memset(head,-1,sizeof(head)); 39 memset(a,0,sizeof(a)); 40 scanf("%d%d",&n,&d); 41 len = 0; 42 for(i = 1; i<n; i++) 43 { 44 scanf("%d%d",&x,&y); 45 add(x,y); 46 } 47 ans = 0; 48 dfs(0,0); 49 printf("%d ",ans); 50 }
并差集：
1 #include<stdio.h> 2 #include<string.h> 3 #include<iostream> 4 #include<algorithm> 5 #define MAXN 110000 6 using namespace std; 7 int pri[MAXN]; 8 int sum,m; 9 int find(int x) 10 { 11 int r=x; 12 while(r!=pri[r]) 13 r=pri[r]; 14 return r; 15 } 16 int num(int a) 17 { 18 int b=0; 19 while(a!=pri[a]) 20 { 21 a=pri[a]; 22 b++; 23 } 24 return b; 25 } 26 void fun() 27 { 28 for(int i=MAXN;i>0;i--) 29 { 30 if(find(i)==0&&num(i)>m)//根节点为0且距离大于m 31 sum++; 32 } 33 } 34 int main() 35 { 36 int n,i,a,b,t; 37 scanf("%d",&t); 38 while(t--) 39 { 40 sum=0; 41 for(i=0;i<MAXN;i++) 42 pri[i]=i; 43 scanf("%d%d",&n,&m); 44 for(i=0;i<n-1;i++) 45 { 46 scanf("%d%d",&a,&b); 47 pri[b]=a;//直接并，不用查 48 } 49 fun(); 50 printf("%d ",sum); 51 } 52 return 0; 53 }