链接：https://ac.nowcoder.com/acm/contest/897/C
来源：牛客网
LaTale
时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
Legend goes that in the heart of ocean, exists a gorgeous island called LaTale, which has n cities. Specially, there is only one way between every two cities.
In other words, n cities construct a tree connected by n-1 edges. Each of the edges has a weight w.
Define d(u, v) the length between city u and city v. Under the condition of u differing from v, please answer how many pairs(u, v) in which d(u, v) can be divisible by 3.Pair(u,v) and pair(v,u) are considered the same.
输入描述:
The first line contains an integer number T, the number of test cases.
For each test case:
The first line contains an integer n(2≤n≤1052≤n≤105), the number of cities.
The following n-1 lines, each contains three integers uiui, vivi, wiwi(1≤ui,vi≤n,1≤wi≤10001≤ui,vi≤n,1≤wi≤1000), the edge of cities.
输出描述:
For each test case print the number of pairs required.
示例1
输出
复制2 6
题意:
给你一个含有n个节点的树，
问有多少对节点u，v，他们的距离dist(u,v)%3==0
思路：
树上dp
我们定义状态 dp[i][0/1/2]
表示第i个节点的子树中，距离第i个节点的距离%3的分别等于0、1、2三种情况的节点个数。
然后根据节点之间的关系转移。
对于一堆节点u，v。对答案的贡献是：
dp[u][j]*dp[v][(3-w-j+3)%3];
j 分别取0 1 2
然后再把底层的节点信息转移给它的父节点即可、
细节见代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define rt return #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d ",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), ' ', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;} inline void getInt(int* p); const int maxn = 100010; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ struct edge { int next; int to; int dis; edge() { } edge(int nn, int dd) { next = nn; dis = dd; } }; edge e[maxn << 1]; ll ans = 0ll; int tot; int head[maxn << 1]; void addedge(int a, int b, int c) { e[tot].to = b; e[tot].dis = c; e[tot].next = head[a]; head[a] = tot++; } void init() { tot = 1; } ll dp[maxn][3]; void dfs(int id, int pre) { dp[id][0] = 1ll; for (int i = head[id]; i != 0; i = e[i].next) { edge x = e[i]; ll w = x.dis; if (x.to != pre) { dfs(x.to, id); ans += dp[id][0] * dp[x.to][(3 - w - 0 + 3) % 3]; ans += dp[id][1] * dp[x.to][(3 - w - 1 + 3) % 3]; ans += dp[id][2] * dp[x.to][(3 - w - 2 + 3) % 3]; dp[id][(0 + w) % 3] += dp[x.to][0]; dp[id][(1 + w) % 3] += dp[x.to][1]; dp[id][(2 + w) % 3] += dp[x.to][2]; } } } int main() { //freopen("D:\common_text\code_stream\in.txt","r",stdin); //freopen("D:\common_text\code_stream\out.txt","w",stdout); int t; gbtb; cin >> t; while (t--) { ans = 0ll; int n; cin >> n; init(); int a, b, c; repd(i, 1, n) { head[i] = 0; dp[i][0] = dp[i][1] = dp[i][2] = 0; } repd(i, 2, n) { cin >> a >> b >> c; c %= 3; addedge(a, b, c); addedge(b, a, c); } dfs(1, 0); cout << ans << endl; } return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ' ' || ch == ' '); if (ch == '-') { *p = -(getchar() - '0'); while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 - ch + '0'; } } else { *p = ch - '0'; while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 + ch - '0'; } } }