题解 | #躲避技能#

差不多的想法，但是用的 topo 实现，把出发点标成 1，目标点标成 -1，然后topo网上搞就行，每条边经过的时候加上这条边的边权乘上当前点的值的绝对值。

保龄了，原因是：

		char s[105];
		scanf ("%s", s);
		len = 0; reverse (s, s + strlen (s));

高精度里面的的这个，考场上写成了

		char s[105];
		scanf ("%s", s);
		len = 0; reverse (s, s + len);

看到翻转但完全没用/ll/ll/ll

代码

# include <bits/stdc++.h>
# define wheneveright signed main
using namespace std;

const int maxn = 300005;
const int maxe = 600009;

const int TT = 10000;
struct INT {
	int len, a[35];
	INT () { len = 0; memset (a, 0, sizeof (a)); }
	void get () {
		char s[105];
		scanf ("%s", s);
		len = 0; reverse (s, s + strlen (s));
		memset(a, 0, sizeof(a));
		int L = strlen (s);
		len = (L + 3) / 4;
		int yu = L % 4 ? L % 4 : 4, cnt = len;
		for (int i = 0; i < yu; i++)a[len] = a[len] * 10 + s[i] - '0';
		for (int i = yu; s[i]; i += 4) {
			int ret = 0;
			for (int j = i; j <= i + 3; j++)ret = ret * 10 + s[j] - '0';
			a[--cnt] = ret;
		}
		return ;
	}
	INT operator + (const INT b) {
		INT c;
		c.len = max(len, b.len);
		for (int i = 1; i <= c.len; i++) {
			c.a[i] += a[i] + b.a[i];
			c.a[i + 1] = c.a[i] / TT;
			c.a[i] %= TT;
		}
		if (c.a[c.len + 1])c.len++;
		return c;
	}
	INT operator * (const INT b) {
		INT c;
		c.len = len + b.len - 1;
		for (int i = 1; i <= len; i++)
			for (int j = 1; j <= b.len; j++) {
				c.a[i + j - 1] += a[i] * b.a[j];
				c.a[i + j] += c.a[i + j - 1] / TT;
				c.a[i + j - 1] %= TT;
			}
		if (c.a[c.len + 1]) c.len++;
		while (c.len > 1 && !c.a[c.len]) c.len--;
		return c;
	}
	void print () {
		printf ("%d", a[len]);
		for (int i = len - 1; i >= 1; i--) printf ("%04d", a[i]);
		printf ("\n");
	}
} vl[maxn], res;

# define int long long

INT change (int x) {
	INT ret; ret.len = 0; memset (ret.a, 0, sizeof ret.a);
	while (x) {
		ret.a[++ret.len] = x % TT;
		x /= TT;
	}
	return ret;
}

struct reader {
	template <typename Type>
	reader & operator >> (Type & ret) {
		int f = 1; ret = 0; char ch = getchar ();
		for (;!isdigit (ch); ch = getchar ()) if (ch == '-') f = -f;
		for (; isdigit (ch); ch = getchar ()) ret = (ret * 10) + ch - '0';
		ret *= f; return * this;
	}
} fin;

int n, m, x, y, z, cnt[maxn];
int lnk[maxn], nxt[maxe], son[maxe], val[maxe], deg[maxn], tot;
void add_e (int x, int y, int z) {
	nxt[++tot] = lnk[x]; lnk[x] = tot; son[tot] = y; val[tot] = z; deg[x]++;
	nxt[++tot] = lnk[y]; lnk[y] = tot; son[tot] = x; val[tot] = z; deg[y]++;
	return ;
}

int a[maxn]; bool vis[maxn];

queue < int > que;
void topo () {
	while (!que.empty ()) que.pop ();
	for (int i = 1; i <= n; i++) if (!--deg[i]) que.push (i);
	while (!que.empty ()) {
		int now = que.front (); que.pop (); vis[now] = true;
		for (int j = lnk[now]; j; j = nxt[j]) {
			if (!vis[son[j]]) cnt[val[j]] += abs (a[now]), a[son[j]] += a[now];
			if (!--deg[son[j]]) que.push (son[j]);
		}
	}
	return ;
}

wheneveright () {
	fin >> n >> m;
	for (int i = 1; i <= m; i++) fin >> x, a[x]++;
	for (int i = 1; i <= m; i++) fin >> x, a[x]--;
	for (int i = 1; i < n; i++) {
		fin >> x >> y; vl[i].get ();
		add_e (x, y, i);
	}
	topo ();
	for (int i = 1; i < n; i++) res = res + (vl[i] * change (cnt[i]));
	res.print ();
	return 0;
}