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  • 洛谷P2634 聪明可可

    还是点分治

    树上问题真有趣ovo,这道题统计模3为0的距离,可以把重心的子树分开统计,也可以一次性统计,然后容斥原理减掉重复的。。
    其他的过程就是点分治的板子啦。

    #include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define full(a, b) memset(a, b, sizeof a)
    using namespace std;
    typedef long long ll;
    inline int lowbit(int x){ return x & (-x); }
    inline int read(){
        int X = 0, w = 0; char ch = 0;
        while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
        while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
        return w ? -X : X;
    }
    inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
    inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
    template<typename T>
    inline T max(T x, T y, T z){ return max(max(x, y), z); }
    template<typename T>
    inline T min(T x, T y, T z){ return min(min(x, y), z); }
    template<typename A, typename B, typename C>
    inline A fpow(A x, B p, C lyd){
        A ans = 1;
        for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
        return ans;
    }
    const int N = 50005;
    int n, cnt, rt, ans, sum, res, head[N], size[N], r[3], cur[3], dis[N];
    bool vis[N];
    struct Edge{ int v, next, w; } edge[N<<1];
    
    void addEdge(int a, int b, int c){
        edge[cnt].v = b, edge[cnt].w = c, edge[cnt].next = head[a], head[a] = cnt ++;
    }
    
    void dfs(int s, int fa){
        int mp = 0;
        size[s] = 1;
        for(int i = head[s]; i != -1; i = edge[i].next){
            int u = edge[i].v;
            if(u == fa || vis[u]) continue;
            dfs(u, s);
            size[s] += size[u];
            mp = max(mp, size[u]);
        }
        mp = max(mp, sum - size[s]);
        if(mp < ans) ans = mp, rt = s;
    }
    
    void getDis(int s, int fa){
        res += r[(3 - dis[s]) % 3];
        cur[dis[s]] ++;
        for(int i = head[s]; i != -1; i = edge[i].next){
            int u = edge[i].v;
            if(u == fa || vis[u]) continue;
            dis[u] = (dis[s] + edge[i].w) % 3;
            getDis(u, s);
        }
    }
    
    void calc(int s){
        r[0] = 1;
        for(int i = head[s]; i != -1; i = edge[i].next){
            int u = edge[i].v;
            if(vis[u]) continue;
            full(cur, 0);
            dis[u] = edge[i].w % 3;
            getDis(u, s);
            for(int j = 0; j < 3; j ++) r[j] += cur[j];
        }
        full(r, 0);
    }
    
    void solve(int s){
        vis[s] = true, calc(s);
        for(int i = head[s]; i != -1; i = edge[i].next){
            int u = edge[i].v;
            if(vis[u]) continue;
            ans = INF, sum = size[u];
            dfs(u, 0), solve(rt);
        }
    }
    
    int main(){
    
        full(head, -1);
        n = read();
        for(int i = 0; i < n - 1; i ++){
            int u = read(), v = read(), c = read() % 3;
            addEdge(u, v, c), addEdge(v, u, c);
        }
        ans = INF, sum = n;
        dfs(1, 0), solve(rt);
        res = (res << 1) + n;
        int t = gcd(res, n * n);
        printf("%d/%d
    ", res / t, n * n / t);
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/onionQAQ/p/10726027.html
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