• BZOJ 1018: [SHOI2008]堵塞的交通traffic


    二次联通门 : BZOJ 1018: [SHOI2008]堵塞的交通traffic

    /*
    
        BZOJ 1018: [SHOI2008]堵塞的交通traffic
    
        麻麻
        这题玩我
    
        这题简直消磨人的意志
        写了一天了
        写一段
        玩一段
    
        直接写不下去
        什么时候恢复一下心情再写
    
    */
    #include <cstdio>
    #include <iostream>
    
    #define rg register
    inline void read (int &n)
    {
        rg char c = getchar ();
        for (n = 0; !isdigit (c); c = getchar ());
        for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
    }
    
    #define Max 200005
    
    struct D 
    {
        int a, b, c, d, e, f;
        D () { a = b = c = d = e = f = 0; }
    
    };
    
    inline D operator + (const D &L, const D &R)
    {
        D n;
        n.d = L.d, n.b = R.b;
        n.a = (L.a & R.a) | (L.e & R.f);
        n.c = (L.c & R.c) | (L.f & R.e);
        n.f = (L.c & R.f) | (L.f & R.a);
        n.e = (L.e & R.c) | (L.a & R.e);
        return n;
    }
    
    
    namespace seg
    {
        int L[Max << 2], R[Max << 2]; D key[Max << 2];
    
        void Build (int n, int l, int r)
        {
            L[n] = l, R[n] = r;
            if (l == r) return ;
            int m = l + r >> 1;
            Build (n << 1, l, m), Build (n << 1 | 1, m + 1, r);
        }
        
        inline void Upsig (int n)
        {
            D &n = key[n];
            n.e = (n.a & n.b) | (n.c & n.d);
            n.f = (n.d & n.d) | (n.c & n.b);
            return ;
        }
    
        void Modi (int n, int p, int t)
        {
            if (L[n] == R[n]) 
            {
                if (t == 1) key[n].a = 1;
                else if (t == 2) key[n].b = 1;
                else if (t == 3) key[n].c = 1;
                else if (t == 4) key[n].d = 1;
    
                if (t == -1) key[n].a = 0;
                else if (t == -2) key[n].b = 0;
                else if (t == -3) key[n].c = 0;
                else if (t == -4) key[n].d = 0;
    
                Upsig (n);    
                return ;
            }
            int m = L[n] + R[n] >> 1;
            if (p <= m) Modi (n << 1, p, t), if (p > m) Modi (n << 1 | 1, p, t);
            key[n] = key[n << 1] + key[n << 1 | 1];
        }
    
    
        D Query (int n, int l, int r)
        {
            if (l <= L[n] && R[n] <= r) return key[n];    
            int m = L[n] + R[n] >> 1; D res;
            if (l <= m) res = Query (n << 1, l, r);
            if (r > m) res = res + Query (n << 1 | 1, l, r);
            return res;    
        }    
    }
    
    int main (int argc, char *argv[])
    {
        int N, M; read (N); rg int i, j;
        int x1, y1, x2, y2; D res; int t;
        for (char type[10]; ; )
        {
            scanf ("%s", type);
            if (type[0] == 'E') break;
            if (type[0] == 'O')
            {
                read (x1), read (y1), read (x2), read (y2);
                if (x1 > x2) std :: swap (x1, x2), std :: swap (y1, y2);
                if (x1 == x2)
                {
                    if (x1 != 1) seg :: Modi (1, x1 - 1, 2), seg :: Modi (1, x1, 4);
                    else seg :: Modi (1, 1, 4);
                }
                else
                {
                    if (y1 == 1) seg :: Modi (1, x1, 1);
                    else seg :: Modi (1, x1, 3);
                }
            }
    
    #define Yes { puts ("Y"); continue; }
    #define No { puts ("N"); continue; }
    
            else if (type[0] == 'Q') // t == 1 
            {
                res = seg :: Query (1, x1, y1 - 1);    
                if (x1 == x2 && y1 == y2) Yes
                if (x1 == x2)
                {
                    if (res.a) Yes
                }        
                if (y1 == y2)
                {
    
                    
                }
            }
            else
            {
    
    
            }
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ZlycerQan/p/8127649.html
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