bzoj4821

线段树

这题真是无聊

把式子拆开，然后可知维护xi,yi,xi^2,xi*yi,重点在于标记下传，当我们进行2号操作时，直接累加进答案和标记即可，进行3号操作时，update时先把自己这层赋值成要改变的值，再清空这层2号标记，每次pushdown把这层的下一层的标记清空，因为下一层被覆盖了，pushdown先执行3号标记，再执行2号标记，因为存在的2号标记肯定在3号标记后打的，否则肯定会被清空，所以2号标记应该在三号标记后。

而下一层的2号标记肯定是在三号标记之前打的，因为标记已经下传。

#include<bits/stdc++.h>
using namespace std;
typedef double ld;
const int N = ;
int n, m;
ld x[N], y[N];
struct node {
    double sumx, sumy, sumxx, sumxy;
    node(ld sumx = , ld sumy = , ld sumxx = , ld sumxy = ) : sumx(sumx), sumy(sumy), sumxx(sumxx), sumxy(sumxy) {}
    void print()
    {
        printf("sumx=%.10f sumy=%.10f sumxx=%.10f sumxy=%.10f\n", sumx, sumy, sumxx, sumxy);
    }
};
struct seg {
    node tree[N << ];
    ld tagx2[N << ], tagy2[N << ], tagx3[N << ], tagy3[N << ];
    bool can1[N << ], can2[N << ];
    ld calc(ld x)
    {
        return (ld)x * (ld)(x + ) * (ld)( * x + ) / ;
    }
    void pushdown(int o, int l, int r)
    {
        int mid = (l + r) >> ;
        if(can1[o])
        {
            tree[o << ].sumx = (ld)(mid - l + ) * tagx3[o] + (ld)(mid + l) * (ld)(mid - l + ) / 2.0;
            tree[o <<  | ].sumx = (ld)(r - mid) * tagx3[o] + (ld)(r + mid + ) * (ld)(r - mid) / 2.0;
            tree[o << ].sumy = (ld)(mid - l + ) * tagy3[o] + (ld)(mid + l) * (ld)(mid - l + ) / 2.0;
            tree[o <<  | ].sumy = (ld)(r - mid) * tagy3[o] + (ld)(r + mid + ) * (ld)(r - mid) / 2.0;
            tree[o << ].sumxx = calc(tagx3[o] + mid) - calc(tagx3[o] + l - );
            tree[o <<  | ].sumxx = calc(tagx3[o] + r) - calc(tagx3[o] + mid);
            tree[o << ].sumxy = (ld)(mid - l + ) * tagx3[o] * tagy3[o] + (ld)(tagx3[o] + tagy3[o]) * (ld)(mid + l) * (ld)(mid - l + ) / 2.0 + calc(mid) - calc(l - );
            tree[o <<  | ].sumxy = (ld)(r - mid) * tagx3[o] * tagy3[o] + (ld)(tagx3[o] + tagy3[o]) * (ld)(r + mid + ) * (ld)(r - mid) / 2.0 + calc(r) - calc(mid);
            tagx3[o << ] = tagx3[o <<  | ] = tagx3[o];
            tagy3[o << ] = tagy3[o <<  | ] = tagy3[o];
            tagx3[o] = tagy3[o] = ;
            tagx2[o << ] = tagx2[o <<  | ] = tagy2[o << ] = tagy2[o <<  | ] = ;
            can2[o << ] = can2[o <<  | ] = ;
            can1[o << ] = can1[o <<  | ] = can1[o];
            can1[o] = ;
        }
        if(can2[o])
        {
            tree[o << ].sumxx += 2.0 * tree[o << ].sumx * tagx2[o] + (ld)(mid - l + ) * tagx2[o] * tagx2[o];
            tree[o <<  | ].sumxx += 2.0 * tree[o <<  | ].sumx * tagx2[o] + (ld)(r - mid) * tagx2[o] * tagx2[o];
            tree[o << ].sumxy += tree[o << ].sumx * tagy2[o] + tree[o << ].sumy * tagx2[o] + (ld)(mid - l + ) * tagx2[o] * tagy2[o];
            tree[o <<  | ].sumxy += tree[o <<  | ].sumx * tagy2[o] + tree[o <<  | ].sumy * tagx2[o] + (ld)(r - mid) * tagx2[o] * tagy2[o];
            tree[o << ].sumx += tagx2[o] * (ld)(mid - l + );
            tree[o <<  | ].sumx += tagx2[o] * (ld)(r - mid);
            tree[o << ].sumy += tagy2[o] * (ld)(mid - l + );
            tree[o <<  | ].sumy += tagy2[o] * (ld)(r - mid);
            tagx2[o << ] += tagx2[o];
            tagx2[o <<  | ] += tagx2[o];
            tagy2[o << ] += tagy2[o];
            tagy2[o <<  | ] += tagy2[o];
            tagx2[o] = ;
            tagy2[o] = ;
            can2[o << ] = can2[o <<  | ] = can2[o];
            can2[o] = ;
        }
    }
    node merge(node B, node C)
    {
        node A;
        A.sumx = B.sumx + C.sumx;
        A.sumy = B.sumy + C.sumy;
        A.sumxx = B.sumxx + C.sumxx;
        A.sumxy = B.sumxy + C.sumxy;
        return A;
    }
    void build(int l, int r, int o)
    {
        if(l == r)
        {
            tree[o] = node(x[l], y[l], x[l] * x[l], x[l] * y[l]);
            return;
        }
        int mid = (l + r) >> ;
        build(l, mid, o << );
        build(mid + , r, o <<  | );
        tree[o] = merge(tree[o << ], tree[o <<  | ]);
    }
    node query(int l, int r, int o, int a, int b)
    {
        if(l > b || r < a) return tree[];
        if(l >= a && r <= b) return tree[o];
        int mid = (l + r) >> ;
        pushdown(o, l, r);
        node tx = query(l, mid, o << , a, b), ty = query(mid + , r, o <<  | , a, b);
        return merge(tx, ty);
    }
    void update2(int l, int r, int o, int a, int b, ld s, ld t)
    {
        if(l > b || r < a) return;
        if(l >= a && r <= b)
        {
            tagx2[o] += s;
            tagy2[o] += t;
            tree[o].sumxx += 2.0 * tree[o].sumx * s + (ld)(r - l + ) * s * s;
            tree[o].sumxy += t * tree[o].sumx + s * tree[o].sumy + (ld)(r - l + ) * s * t;
            tree[o].sumx += (ld)(r - l + ) * s;
            tree[o].sumy += (ld)(r - l + ) * t;
            can2[o] = ;
            return;
        }
        pushdown(o, l, r);
        int mid = (l + r) >> ;
        update2(l, mid, o << , a, b, s, t);
        update2(mid + , r, o <<  | , a, b, s, t);
        tree[o] = merge(tree[o << ], tree[o <<  | ]);
    }
    void update3(int l, int r, int o, int a, int b, ld s, ld t)
    {
        if(l > b || r < a) return;
        if(l >= a && r <= b)
        {
            tagx3[o] = s;
            tagy3[o] = t;
            tagx2[o] = tagy2[o] = ;
            tree[o].sumx = (ld)(r - l + ) * s + (ld)(l + r) * (ld)(r - l + ) / 2.0;
            tree[o].sumy = (ld)(r - l + ) * t + (ld)(l + r) * (ld)(r - l + ) / 2.0;
            tree[o].sumxx = calc(s + r) - calc(s + l - );
            tree[o].sumxy = (ld)(r - l + ) * s * t + ((ld)(l + r) * (ld)(r - l + ) / 2.0) * (ld)(s + t) + calc(r) - calc(l - );
            can1[o] = ;
            can2[o] = ;
            return;
        }
        pushdown(o, l, r);
        int mid = (l + r) >> ;
        update3(l, mid, o << , a, b, s, t);
        update3(mid + , r, o <<  | , a, b, s, t);
        tree[o] = merge(tree[o << ], tree[o <<  | ]);
    }
} t;
int main()
{
    scanf("%d%d", &n, &m);
    for(int i = ; i <= n; ++i)
        scanf("%lf", &x[i]);
    for(int i = ; i <= n; ++i)
        scanf("%lf", &y[i]);
    t.build(, n, );
    for(int i = ; i <= m; ++i)
    {
        int opt, l, r;
        ld S, T;
        scanf("%d", &opt);
        if(opt == )
        {
            scanf("%d%d", &l, &r);
            node o = t.query(, n, , l, r);
            ld ox = o.sumx / (ld)(r - l + ), oy =  o.sumy / (ld)(r - l + );
            ld up = ((ld)(r - l + ) * o.sumxy - o.sumx * o.sumy), down = ((ld)(r - l + ) * o.sumxx - o.sumx * o.sumx), ans = up / down;
            printf("%.10f\n", ans);
        }
        if(opt == )
        {
            scanf("%d%d%lf%lf", &l, &r, &S, &T);
            t.update2(, n, , l, r, S, T);
        }
        if(opt == )
        {
            scanf("%d%d%lf%lf", &l, &r, &S, &T);
            t.update3(, n, , l, r, S, T);
        }
    }
    return ;
}