hdu 4707 Pet 树形DP/DFS 简单题

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本文介绍了林基在发现仓鼠逃逸后，使用地图和陷阱来寻找其可能藏身之处的过程。通过构建学校地图，并结合陷阱的捕获范围，计算出仓鼠可能出现在哪些位置。此过程涉及图论和树形动态规划的知识。

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Pet

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1936 Accepted Submission(s): 942

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

Sample Output

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

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liuyiding

题意:

给你一没有回路的连通图,标记为0~n-1,相邻边的权值为1,找出距离点0 大于m的点的个数.

来一次树形DP，在陷阱捕获范围内的点不算进去就行了

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
#include<utility>
#pragma comment(linker, "/STACK:102400000,102400000")
#define PI 3.1415926535897932384626
#define eps 1e-10
#define sqr(x) ((x)*(x))
#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)
#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)
#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)
#define  lson   num<<1,le,mid
#define rson    num<<1|1,mid+1,ri
#define MID   int mid=(le+ri)>>1
#define zero(x)((x>0? x:-x)<1e-15)
#define mp    make_pair
#define _f     first
#define _s     second

using namespace std;
const int INF =0x3f3f3f3f;
const int maxn=  100000+10  ;
//const int maxm=    ;
//const int INF=    ;
typedef long long ll;
const ll inf =1000000000000000;//1e15;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
//by yskysker123
int e_max,n,D;
int u[maxn*2],v[maxn*2];
int vis[maxn*2];
int fir[maxn];
int nex[maxn*2];
void init()
{
    e_max=0;
    memset(fir,-1,sizeof fir);
    memset(vis,0,sizeof vis);
}

inline void add_edge(int s,int t)
{
    int e=e_max++;
    u[e]=s;
    v[e]=t;
    nex[e]=fir[s];
    fir[s]=e;
}

int  dfs(int x,int step)
{
    int ans=   step<=D? 0:1;
   for(int e=fir[x];~e;e=nex[e])
   {
       int y=v[e];
       if(vis[y])  continue;
       vis[y]=1;
       ans+=dfs(y,step+1);


   }
    return   ans;
}
int main()
{
    int T;int x,y;
    scanf("%d",&T);
    while(T--)
    {
        init();
        scanf("%d%d",&n,&D);
        for(int i=1;i<=n-1;i++)
        {
            scanf("%d%d",&x,&y);
            add_edge(x,y);
            add_edge(y,x);
        }

        vis[0]=1;
       printf("%d\n", dfs( 0 ,0) );
    }


    return 0;
}