Blog of RuSun

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

P1131 [ZJOI2007] 时态同步

P1131 [ZJOI2007] 时态同步

显然这是一棵树。调整越靠根节点的树枝调整的代价越少。先找到最深的叶子节点,再从最小的子树开始,把所有子节点调整到同一深度,再调整子树上面的树枝。

d[x]x 到叶子节点的最长距离,对于每一个 x 的子节点,代价为它们的距离差。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 5e5 + 10, M = 1e6 + 10;
int n, st;
LL ans, d[N];
int idx, hd[N], nxt[M], edg[M], wt[M];
void add(int a, int b, int c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
void dfs(int x, int fa)
{
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] != fa)
{
dfs(edg[i], x);
d[x] = max(d[x], d[edg[i]] + wt[i]);
}
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] != fa)
ans += d[x] - (d[edg[i]] + wt[i]);
}
int main()
{
scanf("%d%d", &n, &st);
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1, a, b, c; i < n; i++)
{
scanf("%d%d%d", &a, &b, &c);
add(a, b, c);
add(b, a, c);
}
dfs(st, 0);
printf("%d", ans);
return 0;
}