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  • 2019 沈阳网络赛 Fish eating fruit

    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
    ", a, b, c);
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/lesning/p/12120923.html
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