• [洛谷P1501][国家集训队]Tree II


    题目大意:给一棵树,有四种操作:

    1. $+;u;v;c:$将路径$u->v$区间加$c$
    2. $-;u_1;v_1;u_2;v_2:$将边$u_1-v_1$切断,改成边$u_2-v_2$,保证数据合法
    3. $*;u;v;c:$将路径$u->v$区间乘$c$
    4. $/;u;v:$询问路径$u->v$区间和

    题解:$LCT$乱搞

    卡点:

    C++ Code:

    #include <cstdio>
    #define maxn 100010
    #define lc(rt) son[rt][0]
    #define rc(rt) son[rt][1]
    const long long mod = 51061;
    int n, q;
    long long V[maxn], s[maxn], tg[maxn], M[maxn], A[maxn];
    int son[maxn][2], fa[maxn], sz[maxn];
    inline void swap(int &a, int &b) {a ^= b ^= a ^= b;}
    inline void swap(int x) {
        swap(lc(x), rc(x));
        tg[lc(x)] ^= 1, tg[rc(x)] ^= 1, tg[x] = 0;
    }
    inline void setmul(int x) {
        long long &tmp = M[x];
        M[lc(x)] = M[lc(x)] * tmp % mod, M[rc(x)] = M[rc(x)] * tmp % mod;
        A[lc(x)] = A[lc(x)] * tmp % mod, A[rc(x)] = A[rc(x)] * tmp % mod;
        V[lc(x)] = V[lc(x)] * tmp % mod, V[rc(x)] = V[rc(x)] * tmp % mod;
        s[lc(x)] = s[lc(x)] * tmp % mod, s[rc(x)] = s[rc(x)] * tmp % mod;
        tmp = 1;
    }
    inline void setadd(int x) {
        long long &tmp = A[x];
        A[lc(x)] = (A[lc(x)] + tmp) % mod, A[rc(x)] = (A[rc(x)] + tmp) % mod;
        V[lc(x)] = (V[lc(x)] + tmp) % mod, V[rc(x)] = (V[rc(x)] + tmp) % mod;
        s[lc(x)] = (s[lc(x)] + sz[lc(x)] * tmp) % mod, s[rc(x)] = (s[rc(x)] + sz[rc(x)] * tmp) % mod;
        tmp = 0;
    }
    inline void pushdown(int x) {
        if (tg[x]) swap(x);
        if (M[x] != 1) setmul(x);
        if (A[x]) setadd(x);
    }
    inline void update(int x) {
        s[x] = (s[lc(x)] + s[rc(x)] + V[x]) % mod;
        sz[x] = sz[lc(x)] + sz[rc(x)] + 1;
    }
    inline int get(int x) {return rc(fa[x]) == x;}
    inline bool isrt(int x) {return lc(fa[x]) != x && rc(fa[x]) != x;}
    inline void rotate(int x) {
        int y = fa[x], z = fa[y], b = get(x);
        if (!isrt(y)) son[z][get(y)] = x;
        fa[son[y][b] = son[x][!b]] = y; son[x][!b] = y;
        fa[y] = x, fa[x] = z;
        update(y), update(x);
    }
    int S[maxn], top;
    inline void splay(int x) {
        S[top = 1] = x;
        for (int y = x; !isrt(y); S[++top] = y = fa[y]);
        for (; top; top--) pushdown(S[top]);
        for (; !isrt(x); rotate(x)) if (!isrt(fa[x])) 
            get(x) ^ get(fa[x]) ? rotate(x) : rotate(fa[x]);
        update(x);
    }
    inline void access(int x) {for (int t = 0; x; rc(x) = t, t = x, x = fa[x]) splay(x);}
    inline void mkrt(int x) {access(x), splay(x), tg[x] ^= 1;}
    inline void link(int x, int y) {mkrt(x), fa[x] = y;}
    inline void split(int x, int y) {mkrt(x), access(y), splay(y);}
    inline void cut(int x, int y) {split(x, y), lc(y) = fa[x] = 0;}
    inline void add(int x, int y, long long num) {
        split(x, y);
        A[y] = (A[y] + num) % mod;
        V[y] = (V[y] + num) % mod;
        s[y] = (s[y] + sz[y] * num) % mod;
    }
    inline void mul(int x, int y, long long num) {
        split(x, y);
        A[y] = A[y] * num % mod;
        V[y] = V[y] * num % mod;
        M[y] = M[y] * num % mod;
        s[y] = s[y] * num % mod;
    }
    inline long long query(int x, int y) {
        split(x, y);
        pushdown(y);
        return s[y];
    }
    
    int main() {
        scanf("%d%d", &n, &q);
        for (int i = 1; i <= n; i++) V[i] = 1, M[i] = 1, s[i] = 1;
        for (int i = 1; i < n; i++) {
            int a, b;
            scanf("%d%d", &a, &b);
            link(a, b);
        }
        while (q --> 0) {
            int x, y;
            long long z;
            char op[10];
            scanf("%s%d%d", op, &x, &y);
            switch (*op) {
                case '+': scanf("%lld", &z), add(x, y, z); break;
                case '-': cut(x, y), scanf("%d%d", &x, &y), link(x, y); break;
                case '*': scanf("%lld", &z), mul(x, y, z); break;
                case '/': printf("%lld
    ", query(x, y));
            }
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Memory-of-winter/p/9657097.html
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