Codeforces Round #474 E. Alternating Tree

E. Alternating Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Given a tree with $\mathrm{nn nodes\; numbered\; from 11 to nn.\; Each\; node ii has\; an\; associated\; value ViVi.}$

If the simple path from ${\mathrm{u1u1 to umum consists\; of mm nodes\; namely u1\to u2\to u3\to \dots um-1\to umu1\to u2\to u3\to \dots um-1\to um,\; then\; its\; alternating\; function A(u1,um)A(u1,um)is\; defined\; as A(u1,um)=m\sum i=1(-1)i+1\cdot VuiA(u1,um)=\sum i=1m(-1)i+1\cdot Vui.\; A\; path\; can\; also\; have 00 edges,\; i.e. u1=umu1=um.}}^{}$

Compute the sum of alternating functions of all unique simple paths. Note that the paths are directed: two paths are considered different if the starting vertices differ or the ending vertices differ. The answer may be large so compute it modulo ${\mathrm{109+7109+7.}}^{}$

Input

The first line contains an integer $\mathrm{nn (2\le n\le 2\cdot 105)(2\le n\le 2\cdot 105) —\; the\; number\; of\; vertices\; in\; the\; tree.}$

The second line contains $\mathrm{nn space-separated\; integers V1,V2,\dots ,VnV1,V2,\dots ,Vn (-109\le Vi\le 109-109\le Vi\le 109)\; —\; values\; of\; the\; nodes.}$

The next $\mathrm{n-1n-1 lines\; each\; contain\; two\; space-separated\; integers uu and vv (1\le u,v\le n,u\ne v)(1\le u,v\le \hspace{0.17em}n,u\ne v) denoting\; an\; edge\; between\; vertices uuand vv.\; It\; is\; guaranteed\; that\; the\; given\; graph\; is\; a\; tree.}$

Output

Print the total sum of alternating functions of all unique simple paths modulo ${\mathrm{109+7109+7.}}^{}$

Examples

input

4
-4 1 5 -2
1 2
1 3
1 4

output

40

input

8
-2 6 -4 -4 -9 -3 -7 23
8 2
2 3
1 4
6 5
7 6
4 7
5 8

output

4

思路：只需求奇数长度路径权值和即可，因为偶数长度来回跑会消掉。树上dp，每个点维护从以此点为起点，终点在子树内的奇长路径数，偶长路径数，奇长路径权值和，偶长路径权值和即可。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
*/

int n;
ll V[200015], sume[200015], sumo[200015], cnte[200015], cnto[200015], sum;
void mod(ll &x) {
    x = x % MOD;
    if (x < 0) x += MOD;
}
vector<int> E[200015];
void dfs(int x, int p) {
    ll cur_sume = 0, cur_sumo = 0, cur_cnte = 0, cur_cnto = 0;
    for (int i = 0; i < E[x].size(); ++i) {
        int y = E[x][i];
        if (y == p) continue;
        dfs(y, x);
        cur_sume += sume[y];
        cur_sumo += sumo[y];
        cur_cnte += cnte[y];
        cur_cnto += cnto[y];
    }
    mod(cur_cnte), mod(cur_cnto), mod(cur_sume), mod(cur_sumo);
    for (int i = 0; i < E[x].size(); ++i) {
        int y = E[x][i];
        if (y == p) continue;
        ll cnte_y = cnte[y], cnte_other = cur_cnte - cnte_y;
        ll cnto_y = cnto[y], cnto_other = cur_cnto - cnto_y;
        ll sume_y = sume[y], sume_other = cur_sume - sume_y;
        ll sumo_y = sumo[y], sumo_other = cur_sumo - sumo_y;
        sum += cnte_y * sume_other + cnte_other * sume_y + cnte_other * cnte_y % MOD * V[x];
        sum += cnto_y * sumo_other + cnto_other * sumo_y - cnto_other * cnto_y % MOD * V[x];
        mod(sum);
    }
    cnto[x] = 1 + cur_cnte;
    cnte[x] = cur_cnto;
    sumo[x] = cur_cnte * V[x] + cur_sume;
    sume[x] = -cur_cnto * V[x] + cur_sumo;
    sum += 2 * sumo[x];
    sumo[x] += V[x];
    mod(cnto[x]), mod(cnte[x]), mod(sumo[x]), mod(sume[x]), mod(sum);
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%lld", &V[i]);
        sum += V[i];
    }
    for (int i = 1; i < n; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        E[u].pb(v);
        E[v].pb(u);
    }
    dfs(1, 0);
    printf("%lld
", sum);
    return 0;
}
/*
4
4 1 5 2
1 2
1 3
1 4
*/