Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P3066 [USACO12DEC]Running Away From the Barn G

P3066 [USACO12DEC]Running Away From the Barn G

维护一个栈记录从根节点到当前节点的所有节点。维护一个指针总是保证从该位置到当前节点的距离小于 $t$ ,则当前节点对两点之间的所有节点都有贡献,用树上差分统计答案。

查看代码
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#include <cstdio>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10, M = 2e5 + 10;
int n, p[N], c[N], f[N];
int idx, hd[N], nxt[M], edg[M];
LL s, w[N];
int top, stk[N];
void dfs(int x, int t, LL d)
{
stk[++top] = x;
while (d > s)
d -= w[stk[++t]];
c[x]++, c[p[stk[t]]]--;
for (int i = hd[x]; ~i; i = nxt[i])
dfs(edg[i], t, d + w[edg[i]]);
top--;
}
int cal(int x)
{
int res = c[x];
for (int i = hd[x]; ~i; i = nxt[i])
res += cal(edg[i]);
return f[x] = res;
}
void add(int a, int b)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
}
int main()
{
scanf("%d%lld", &n, &s);
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 2; i <= n; i++)
{
scanf("%d%lld", &p[i], &w[i]);
add(p[i], i);
}
dfs(1, 1, 0);
cal(1);
for (int i = 1; i <= n; i++)
printf("%d\n", f[i]);
return 0;
}