C. Uncle Bogdan and Country Happiness solution _Solution

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C. Uncle Bogdan and Country Happinesstime limit per test 2 secondsmemory limit per test 256 megabytesinput standard inputoutput standard outputUncle Bogdan is in captain Flint‘s crew for a long time and sometimes gets nostalgic for his homeland. Today he told you how his country introduced a happiness index.
There are nn cities and n− 1n− 1 undirected roads connecting pairs of cities. Citizens of any city can reach any other city traveling by these roads. Cities are numbered from 11 to nn and the city 11 is a capital. In other words, the country has a tree structure.
There are mm citizens living in the country. A pipi people live in the ii-th city but all of them are working in the capital. At evening all citizens return to their home cities using the shortest paths.
Every person has its own mood: somebody leaves his workplace in good mood but somebody are already in bad mood. Moreover any person can ruin his mood on the way to the hometown. If person is in bad mood he won‘t improve it.
Happiness detectors are installed in each city to monitor the happiness of each person who visits the city. The detector in the ii-th city calculates a happiness index hihi as the number of people in good mood minus the number of people in bad mood. Let‘s say for the simplicity that mood of a person doesn‘t change inside the city.
Happiness detector is still in development, so there is a probability of a mistake in judging a person‘s happiness. One late evening, when all citizens successfully returned home, the government asked uncle Bogdan (the best programmer of the country) to check the correctness of the collected happiness indexes.
【C. Uncle Bogdan and Country Happiness solution】Uncle Bogdan successfully solved the problem. Can you do the same?
More formally, You need to check: "Is it possible that, after all people return home, for each city ii the happiness index will be equal exactly to hihi".
InputThe first line contains a single integer tt (1≤ t≤ 100001≤ t≤ 10000) — the number of test cases.
The first line of each test case contains two integers nn and mm (1≤ n≤ 1051≤ n≤ 105; 0≤ m≤ 1090≤ m≤ 109) — the number of cities and citizens.
The second line of each test case contains nn integers p1,p2,… ,pnp1,p2,… ,pn (0≤ pi≤ m0≤ pi≤ m; p1+p2+… +pn=mp1+p2+… +pn=m), where pipi is the number of people living in the ii-th city.
The third line contains nn integers h1,h2,… ,hnh1,h2,… ,hn (− 109≤ hi≤ 109− 109≤ hi≤ 109), where hihi is the calculated happiness index of the ii-th city.
Next n− 1n− 1 lines contain description of the roads, one per line. Each line contains two integers xixi and yiyi (1≤ xi,yi≤ n1≤ xi,yi≤ n; xi≠ yixi≠ yi), where xixi and yiyi are cities connected by the ii-th road.
It‘s guaranteed that the sum of nn from all test cases doesn‘t exceed 2⋅ 1052⋅ 105.
OutputFor each test case, print YES, if the collected data is correct, or NO — otherwise. You can print characters in YES or NO in any case.
Examplesinput Copy

2 7 4 1 0 1 1 0 1 0 4 0 0 -1 0 -1 0 1 2 1 3 1 4 3 5 3 6 3 7 5 11 1 2 5 2 1 -11 -2 -6 -2 -1 1 2 1 3 1 4 3 5

output Copy

YES YES

input Copy

2 4 4 1 1 1 1 4 1 -3 -1 1 2 1 3 1 4 3 13 3 3 7 13 1 4 1 2 1 3

output Copy

NO NO
solution：

#include< iostream> #include< vector> using namespace std; const int N = 1e5+5; vector< int> buf[N]; int a[N],p[N],h[N],g[N]; bool flag = true; void dfs(int v, int ancestor = -1) { a[v] = p[v]; int sum = 0,to,i; for(i=0; i< buf[v].size(); i++){ to=buf[v][i]; if(to == ancestor) continue; dfs(to, v); a[v] += a[to]; sum += g[to]; //统计now节点为根的子树的总人数 } g[v] = (a[v]+h[v])/2; if((a[v]+h[v])%2 != 0||g[v] < 0 || g[v] > a[v]||sum > g[v]) flag = false; /*1.说明不是整数 2.说明快乐人数比总人数还多 3.说明经过now城市后的快乐人数竟然比走的更远的城市的快乐人数更少*/ } int main(int argc, const char** argv) { ios::sync_with_stdio(false); cin.tie(); int t; cin> > t; while(t--){ int n,m; cin> > n> > m; for(int i = 1; i < = n; ++i) cin> > p[i]; for(int i = 1; i < = n; ++i) cin> > h[i]; for(int i = 0; i < n-1; ++i){ int a,b; cin> > a> > b; buf[a].push_back(b); buf[b].push_back(a); } dfs(1); if (flag) cout < < "YES" < < endl; else cout < < "NO" < < endl; flag = true; for(int i = 1; i < =n; ++i) buf[i].clear(); } return 0; }