D. Valid Sets
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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
Copy
1 4
2 1 3 2
1 2
1 3
3 4
Output
8
Input
Copy
0 3
1 2 3
1 2
2 3
Output
3
Input
Copy
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.
题解
直接统计不好统计，所以考虑把方案分类。
按照最大值点对方案分类，枚举一个最大值点，让这一个点在方案中，这样就确定了方案能包含的连通块。
设(dp[i])表示一定包含i这个点且i这个点是方案中权值最大的点的方案，有
这样会计算重。因为最大值点所确定的连通块里，可能有多个与最大值点权值相同的点，每个点都会算一遍。
只算一个，然后乘以连通的相同值点的个数？
不！别忘了我们(dp)的时候，要求选定的点必须在方案中。必定包含不同的相同最大权值点的方案并不是一一对应的。
怎么办？
继续把方案分类。
在这个包含多个最大值点的极大连通块里，设有(k)个相同的最大值点。
设最大的相同点编号为(x)，(dp)必定包含点x的集合就行了。
这就相当于，我们只走比他编号小的相同权值的点。
两次使用最大值分类方案的神题Orz
被long long 和 MOD卡了两次。。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
inline long long max(long long a, long long b){return a > b ? a : b;}
inline long long min(long long a, long long b){return a < b ? a : b;}
inline long long abs(long long x){return x < 0 ? -x : x;}
inline void swap(long long &x, long long &y){long long tmp = x;x = y;y = tmp;}
inline void read(long long &x)
{
x = 0;char ch = getchar(), c = ch;
while(ch < '0' || ch > '9') c = ch, ch = getchar();
while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
if(c == '-') x = -x;
}
const long long INF = 0x3f3f3f3f;
const long long MAXN = 2000 + 10;
const long long MOD = 1000000007;
struct Edge
{
long long u,v,nxt;
Edge(long long _u, long long _v, long long _nxt){u = _u, v =_v, nxt = _nxt;}
Edge(){}
}edge[MAXN << 1];
long long head[MAXN], cnt, n, val[MAXN], d, dp[MAXN], ans;
inline void insert(long long a, long long b)
{
edge[++ cnt] = Edge(a, b, head[a]), head[a] = cnt;
}
void dfs(long long x, long long pre, long long root)
{
dp[x] = 1;
for(long long pos = head[x];pos;pos = edge[pos].nxt)
{
long long v = edge[pos].v;
if(v == pre || val[root] - val[v] < 0 || val[root] - val[v] > d || (val[v] == val[root] && v > root)) continue;
dfs(v, x, root);
dp[x] *= (dp[v] + 1), dp[x] %= MOD;
}
}
int main()
{
read(d), read(n);
for(long long i = 1;i <= n;++ i) read(val[i]);
for(long long i = 1;i < n;++ i)
{
long long tmp1, tmp2;
read(tmp1), read(tmp2);
insert(tmp1, tmp2), insert(tmp2, tmp1);
}
for(long long i = 1;i <= n;++ i)
{
memset(dp, 0, sizeof(dp));
dfs(i, -1, i);
ans += dp[i] % MOD;
if(ans >= MOD) ans -= MOD;
}
printf("%I64d", ans);
return 0;
}