这题看了三个月,终于过了,第一次看的时候没学树形DP,想用点分治但是不会
后来学了二次扫描,就有点想法了。。。。
这东西也真就玄学了吧。。。
#include<iostream> #include<cstring> #include<vector> #include<algorithm> using namespace std; typedef long long ll; const int maxn = 1e5 + 7; const ll mod = 1e9 + 7; struct Node { int p; ll len; Node(int _p, ll _len) :p(_p), len(_len) {} }; int n; ll dp[maxn][5]; ll cnt[maxn][6]; vector<Node>G[maxn]; void insert(int be, int en, ll len) { G[be].push_back(Node(en, len)); } int dfs2(int x, int fa) { for (int i = 0; i < G[x].size(); i++) { int p = G[x][i].p; ll len = G[x][i].len; if (p == fa) continue; dfs2(p, x); for (int a = 0; a < 3; a++) { dp[x][(a + len) % 3] += (dp[p][a] + cnt[p][a] * len) % mod; cnt[x][(a + len) % 3] += cnt[p][a]; dp[x][(a + len) % 3] %= mod; } dp[x][len % 3] += len; dp[x][len % 3] %= mod; cnt[x][len % 3] ++; } return 0; } ll ans[10]; ll son[10]; int dfs(int x, int fa) { for (int i = 0; i < G[x].size(); i++) { int p = G[x][i].p; ll len = G[x][i].len; if (p == fa) continue; for (int a = 0; a < 3; a++) { ans[(a + len) % 3] = (dp[x][(a + len) % 3] - (cnt[p][a] * len + dp[p][a])) % mod; ans[(a + len) % 3] += mod; ans[(a + len) % 3] %= mod; son[(a + len) % 3] = cnt[x][(a + len) % 3] - cnt[p][a]; } son[len % 3]--; ans[len % 3] = (ans[len % 3] - len + mod) % mod; //删除了多的边 for (int a = 0; a < 3; a++) { dp[p][(a + len) % 3] += (ans[a] + son[a] * len) % mod; dp[p][(a + len) % 3] %= mod; cnt[p][(a + len) % 3] += son[a]; } cnt[p][len % 3]++; dp[p][len % 3] += len; dp[p][len % 3] %= mod; dfs(p, x); } return 0; } int main() { while (~scanf("%d", &n)) { for (int i = 0; i <= n; i++) G[i].clear(); memset(dp, 0, sizeof(dp)); memset(cnt, 0, sizeof(cnt)); int be, en; ll len; for (int i = 1; i < n; i++) { scanf("%d %d %lld", &be, &en, &len); insert(be, en, len); insert(en, be, len); } dfs2(0, -1); dfs(0, -1); ll a = 0, b = 0, c = 0; for (int i = 0; i < n; i++) { a = (a + dp[i][0]) % mod; b = (b + dp[i][1]) % mod; c = (c + dp[i][2]) % mod; } printf("%lld %lld %lld\n", a, b, c); } return 0; }