codeforces 486d Valid Sets

codeforces 486d Valid Sets

​​ As you know, an undirected connected graph with n nodes and n - 1 edges is called a tree. You are given an integer d and a tree consisting of n nodes. Each node i has a value ai associated with it. We call a set S of tree nodes valid if following conditions are satisfied: S is non-empty. S is connected. In other words, if nodes u and v are in S, then all nodes lying on the simple path between u and v should also be presented in S. . Your task is to count the number of valid sets. Since the result can be very large, you must print its remainder modulo 1000000007 (109 + 7). Input The first line contains two space-separated integers d (0 ≤ d ≤ 2000) and n (1 ≤ n ≤ 2000). The second line contains n space-separated positive integers a1, a2, …, an(1 ≤ ai ≤ 2000). Then the next n - 1 line each contain pair of integers u and v (1 ≤ u, v ≤ n) denoting that there is an edge between u and v. It is guaranteed that these edges form a tree. Output Print the number of valid sets modulo 1000000007. Examples Input 1 4 2 1 3 2 1 2 1 3 3 4 Output 8 Input 0 3 1 2 3 1 2 2 3 Output 3 Input 4 8 7 8 7 5 4 6 4 10 1 6 1 2 5 8 1 3 3 5 6 7 3 4 Output 41 Note In the first sample, there are exactly 8 valid sets: {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {3, 4} and {1, 3, 4}. Set {1, 2, 3, 4} is not valid, because the third condition isn’t satisfied. Set {1, 4} satisfies the third condition, but conflicts with the second condition.

题目要求我们求一个连续的区间 满足这个区间内最大值和最小值小于等于给定的d 然后去求 这样的区间的数量 考虑一种树形dp的方法 设定dp[x] 为认为我当前x号的值是最大的然后去求一下可以到哪里 我们去dp[y] y这个点可以选 那么就递归dfs去搜索y这个点的子树求出结果 如果不选y这个点 那么y的子树也一定不会被选 所以直接略过就好 但是我们可能有重复的 怎么办 规定一下方向保证只算一次 然后统计答案自己画画图就好

#include#define N 2200#define mod 1000000007#define ll long longinline char gc(){ static char now[1<<16],*S,*T; if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;} return *S++;}inline int read(){ int x=0;char ch=gc(); while (ch<'0'||ch>'9') ch=gc(); while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();} return x;}int h[N],w[N],fa[N],n,d,num,dp[N];struct node{ int y,next;}data[N<<1];void dfs(int x,int top){ dp[x]=1; for (int i=h[x];i;i=data[i].next){ int y=data[i].y;if (fa[x]==y) continue;int tmp=w[top]-d;if (tmp<0) tmp=0; if (w[y]w[top]) continue;fa[y]=x; dfs(y,top);dp[x]=(ll)dp[x]*(dp[y]+1)%mod; fa[y]=0; }}int main(){ freopen("cf.in","r",stdin); d=read();n=read(); for (int i=1;i<=n;++i) w[i]=read(); for (int i=1;i

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