这是自己一个私题的代码

但不明白为啥出现了一个问题：

在读入样例时，用cin读入type数组能得到正解

而用快读读入type数组得到的结果是错的

输入样例：

10 11
0 1 0 1 0 1 0 1 0 1
1 2
1 3
1 4
2 5
2 6
3 7
4 8
5 9
9 10
5 6
2 7

输出样例：

6050

代码

/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
using namespace std;
const int MAXN = 1e5+5;
const int MAXM = 2e5+5;
const int mod = 1000000007;
struct edge{
	int to, w, nxt;
}e[MAXM];
int head[MAXN], num_edge;
int n, m;
int dis[MAXN], fath[MAXN], f[MAXN], siz[MAXN];
bool type[MAXN];

int read(){
	int w = 1, s = 0;
	char ch = getchar();
	while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
	while(ch >= '0' && ch <= '9') s = s* 10 + ch - '0', ch = getchar();
	return s * w;
}
//bool read1(){
//	bool s = 0;
//	char ch = getchar();
//	while(ch >= '0' && ch <= '9') s = s* 10 + ch - '0', ch = getchar();
//	return s;
//}

void add(int from, int to, int w){
	e[++num_edge].to = to;
	e[num_edge].w = w;
	e[num_edge].nxt = head[from];
	head[from] = num_edge;
} 

int find(int x){return fath[x] == x ? x : fath[x] = find(fath[x]); }

void dfs(int x, int fa){
	for(int i = head[x]; i; i = e[i].nxt){
		int v = e[i].to;
		if(v == fa) continue;
		dis[v] = (dis[x] + e[i].w) % mod;
		dfs(v, x), siz[x] += siz[v];
	}
}

void dfs2(int x, int fa){
	for(int i = head[x]; i; i = e[i].nxt){
		int v = e[i].to;
		if(v == fa) continue;
		int jia = ((siz[1] - siz[v]) * e[i].w % mod + mod) % mod;
		int jian = siz[v] * e[i].w % mod;
		f[v] = ((f[x] + jia - jian) % mod + mod) % mod;
		dfs2(v, x); 
	}
}

int main()
{
	n = read(), m = read();
	bool flag, se;
	se = read();
	if(se) flag = 1, type[1] = se ^ 1;
	else type[1] = 0;
	for(int i = 2; i <= n; ++i){
//		type[i] = (bool)read();
		cin>>type[i];
		if(flag) type[i] ^= 1;
		siz[i] = type[i];//顺便处理在以i为根的子树中，与1节点不同的点的个数 
	}
	for(int i = 1; i <= n; ++i) fath[i] = i;
	int x = 2, cnt = 0;
	for(int i = 1, u, v; i <= m; ++i){
		u = read(), v = read();
		int uf = find(u), vf = find(v);
		if(uf != vf){
			fath[uf] = vf;
			add(u, v, x), add(v, u, x);
			if(++cnt == n - 1) break;
		}
		x = x * 2 % mod;
	}
	dfs(1, 1);
	for(int i = 1; i <= n; ++i){
		if(type[i]) f[1] = (f[1] + dis[i]) % mod;//先暴力把1与其他点的权值和加起来 
	} 
	dfs2(1, 1);
	int ans = 0;
	for(int i = 1; i <= n; ++i){
		if(!type[i]) ans = (ans + f[i]) % mod;
	}
	printf("%d", ans);
	return 0;
}