题目链接：点击打开链接
题意：给定常数d 和 n，表示n个点的树，每个点有点权
问：有多少个点集使得点集中的点两两可达且最大点权-最小点权<=d
思路:
对于每个点，计算包含这个点且这个点作为点集中的最小点权时的个数，枚举每个点。
import java.io.PrintWriter;import java.text.DecimalFormat;import java.util.ArrayDeque;import java.util.ArrayList;import java.util.Arrays;import java.util.Collections;import java.util.Comparator;import java.util.Deque;import java.util.HashMap;import java.util.Iterator;import java.util.LinkedList;import java.util.Map;import java.util.PriorityQueue;import java.util.Scanner;import java.util.Stack;import java.util.TreeMap;import java.util.TreeSet;import java.util.Queue;public class Main {int n, d;int[] a = new int[N];int ad(int x, int y){x += y;if(x>=mod)x %= mod;return x;}int mul(int x, int y){long xx = (long)x*(long)y;if(xx>=mod) xx %= mod;return (int)xx;}int dfs(int u, int fa, int dep){int ans = 1;for(int i = head[u]; i!=-1; i = edge[i].nex){int v = edge[i].to;if((v == fa)||(a[v]>a[dep]||(a[v]==a[dep]&&v>dep))||(a[dep]-a[v]>d))continue;ans = mul(ans, dfs(v, u, dep)+1);}return ans;}void input(){d = cin.nextInt(); n = cin.nextInt();for(int i = 1; i <= n; i++)a[i] = cin.nextInt();init_edge();for(int i = 1, u, v; i < n; i++){u = cin.nextInt(); v = cin.nextInt();add(u, v); add(v, u);}}void work(){input();int ans = 0;for(int i = 1; i <= n; i++)ans = ad(ans, dfs(i, i, i));out.println(ans%mod);}Main() {cin = new Scanner(System.in);out = new PrintWriter(System.out);}public static void main(String[] args) {Main e = new Main();e.work();out.close(); }public Scanner cin;public static PrintWriter out;static int N = 2005;static int M = 2005;DecimalFormat df=new DecimalFormat("0.0000");static int inf = (int) 1e9 + 7;static long inf64 = (long) 1e18;static double eps = 1e-8;static double Pi = Math.PI;static int mod = 1000000007 ;class Edge{int from, to, nex;Edge(){}Edge(int from, int to, int nex){this.from = from;this.to = to;this.nex = nex;}}Edge[] edge = new Edge[M<<1];int[] head = new int[N];int edgenum;void init_edge(){for(int i = 0; i < N; i++)head[i] = -1; edgenum = 0;}void add(int u, int v){edge[edgenum] = new Edge(u, v, head[u]);head[u] = edgenum++;}int Pow(int x, int y) {int ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;y >>= 1;x = x * x;}return ans;}double Pow(double x, int y) {double ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;y >>= 1;x = x * x;}return ans;}int Pow_Mod(int x, int y, int mod) {int ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;ans %= mod;y >>= 1;x = x * x;x %= mod;}return ans;}long Pow(long x, long y) {long ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;y >>= 1;x = x * x;}return ans;}long Pow_Mod(long x, long y, long mod) {long ans = 1;while (y > 0) {if ((y & 1) > 0)ans *= x;ans %= mod;y >>= 1;x = x * x;x %= mod;}return ans;}int gcd(int x, int y){if(x>y){int tmp = x; x = y; y = tmp;}while(x>0){y %= x;int tmp = x; x = y; y = tmp;}return y;}int max(int x, int y) {return x > y ? x : y;}int min(int x, int y) {return x < y ? x : y;}double max(double x, double y) {return x > y ? x : y;}double min(double x, double y) {return x < y ? x : y;}long max(long x, long y) {return x > y ? x : y;}long min(long x, long y) {return x < y ? x : y;}int abs(int x) {return x > 0 ? x : -x;}double abs(double x) {return x > 0 ? x : -x;}long abs(long x) {return x > 0 ? x : -x;}boolean zero(double x) {return abs(x) < eps;}double sin(double x){return Math.sin(x);}double cos(double x){return Math.cos(x);}double tan(double x){return Math.tan(x);}double sqrt(double x){return Math.sqrt(x);}}
，有山就有路，有河就能渡。
