BZOJ 2435：[Noi2011]道路修建（树型DP）
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 }
