• BZOJ4568: [Scoi2016]幸运数字(线性基 倍增)


    题意

    题目链接

    Sol

    线性基是可以合并的

    倍增维护一下

    然后就做完了??

    喵喵喵?

    // luogu-judger-enable-o2
    #include<bits/stdc++.h>
    #define LL long long 
    using namespace std;
    const int MAXN = 2e4 + 10, B = 60;
    inline LL read() {
        char c = getchar(); LL x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N, Q, g[MAXN][16], dep[MAXN];
    LL a[MAXN];
    vector<int> v[MAXN];
    struct Node {
        LL P[62];
        Node() {
            memset(P, 0, sizeof(P));
        }
        void insert(LL x) {
            for(int i = B; i >= 0; i--) {
                if((!(x >> i & 1))) continue;
                if(P[i]) {x ^= P[i]; continue;}
                P[i] = x; break;
            }
        }
        LL QueryMax() {
            LL ans = 0;
            for(int i = B; i >= 0; i--) ans = max(ans, ans ^ P[i]);
            return ans;
        }
        Node operator + (const Node &rhs) const {
            Node ans; 
            for(int i = 0; i <= B; i++) ans.P[i] = this -> P[i];
            for(int i = 0; i <= B; i++) if(rhs.P[i]) ans.insert(rhs.P[i]);
            return ans; 
        }
    }f[MAXN][16];
    void dfs(int x, int fa) {
        dep[x] = dep[fa] + 1;
        g[x][0] = fa;
        f[x][0].insert(a[x]); f[x][0].insert(a[fa]);
        for(int i = 0; i < v[x].size(); i++) {
            int to = v[x][i];
            if(to == fa) continue;
            dfs(to, x);
        }
    }
    void Pre() {
        for(int j = 1; j <= 15; j++)
            for(int i = 1; i <= N; i++)
                g[i][j] = g[g[i][j - 1]][j - 1], 
                f[i][j] = f[i][j - 1] + f[g[i][j - 1]][j - 1];
    }
    LL Query(int x, int y) {
        Node base; base.insert(a[x]); base.insert(a[y]);
        if(dep[x] < dep[y]) swap(x, y);
        for(int i = 15; i >= 0; i--)
            if(dep[g[x][i]] >= dep[y])
                base = base + f[x][i], x = g[x][i];
        if(x == y) return base.QueryMax();
        for(int i = 15; i >= 0; i--)
            if(g[x][i] != g[y][i]) {
                base = base + f[x][i];
                base = base + f[y][i];
                x = g[x][i], y = g[y][i];
            }
        base = base + f[x][0]; base = base + f[y][0];
        return base.QueryMax();
    }
    int main() {
    #ifndef ONLINE_JUDGE
        //freopen("a.in", "r", stdin); freopen("b.out", "w", stdout);
    #endif
        N = read(); Q = read();
        for(int i = 1; i <= N; i++) a[i] = read();
        for(int i = 1; i <= N - 1; i++) {
            int x = read(), y = read();
            v[x].push_back(y); v[y].push_back(x);
        }
        dfs(1, 0);
        Pre();
        while(Q--) {
            int x = read(), y = read();
            printf("%lld
    ", Query(x, y));
        }
        return 0;
    }
    /*
    8 4
    45654 251 321 54687 321321 654 6321432 5646
    1 2
    2 3
    2 4
    1 5
    4 6 
    6 7
    5 8
    
    7 8
    7 5
    6 8
    4 1
    */
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10059651.html
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