BZOJ 2435：[Noi2011]道路修建（树型DP）

Posted 2020-08-30 Shadowdsp

http://www.lydsy.com/JudgeOnline/problem.php?id=2435

题意：中文题意。

思路：很简单的树形DP，sz记录儿子有多少个和cur记录走的哪条弧，然后直接算就可以了。（时间有点紧）。

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 #define N 1000010
 7 struct Edge {
 8     int v, w, nxt;
 9 } edge[N*2];
10 int head[N], tot, cur[N]; long long sz[N];
11 
12 void Add(int u, int v, int w) {
13     edge[tot] = (Edge) {v, w, head[u]}; head[u] = tot++;
14     edge[tot] = (Edge) {u, w, head[v]}; head[v] = tot++;
15 }
16 
17 void DFS(int u, int fa) {
18     sz[u] = 1;
19     for(int i = head[u]; ~i; i = edge[i].nxt) {
20         int v = edge[i].v;
21         if(v == fa) continue;
22         cur[v] = i;
23         DFS(v, u);
24         sz[u] += sz[v];
25     }
26 }
27 
28 int main() {
29     int n;
30     while(~scanf("%d", &n)) {
31         tot = 0;
32         memset(head, -1, sizeof(head));
33         memset(sz, 0, sizeof(sz));
34         for(int i = 1; i < n; i++) {
35             int u, v, w;
36             scanf("%d%d%d", &u, &v, &w);
37             Add(u, v, w);
38         }
39         DFS(1, -1);
40         long long ans = 0;
41         for(int i = 2; i <= n; i++) {
42             ans += (long long)edge[cur[i]].w * abs(n - sz[i] - sz[i]);
43         }
44         printf("%lld\n", ans);
45     }
46     return 0;
47 }