BZOJ 4127 Abs 解题报告

这个题感觉很厉害的样子。。

首先我们注意到一点：每次加的 $d$ 都是非负的。

那么就说明一个数只可能从负数变成非负数并且只会变一次。

所以我们就可以暴力地去改变一个数的正负情况。

然后我们就可以用树链剖分，维护一下区间的最大负数和负数的个数就可以了。

时间复杂度 $O(n\log^2 n)$，空间复杂度 $O(n)$。

 #include <cstdio>
 typedef long long LL;
 #define N 262144 + 5
 #define INF 1234567890987654321LL

 int n, m, tot, cnt, A[N], Head[N], Root[N], Fa[N], Ad[N], Pos[N], Num[N], Dfn[N], Len[N], Size[N], Son[N], q[N];

 struct Edge
 {
     int next, node;
 }E[N];

 struct Segment_Tree
 {
     int l, r, neg_cnt;
     LL sum, delta, neg_Max;
 }h[N];

 inline void addedge(int u, int v)
 {
     E[++ tot].next = Head[u], Head[u] = tot;
     E[tot].node = v;
     E[++ tot].next = Head[v], Head[v] = tot;
     E[tot].node = u;
 }

 inline LL Abs(LL x)
 {
     return x >  ? x : -x;
 }

 inline void dfs(int z)
 {
     Size[z] = ;
     for (int i = Head[z]; i; i = E[i].next)
     {
         int d = E[i].node;
         if (d == Fa[z]) continue ;
         Fa[d] = z;
         dfs(d);
         Size[z] += Size[d];
     }
     for (int i = Head[z]; i; i = E[i].next)
     {
         int d = E[i].node;
         if (d == Fa[z]) continue ;
         if (!Son[z] || Size[d] > Size[Son[z]])
             Son[z] = d;
     }
 }

 inline void update(int x)
 {
     h[x].sum = h[h[x].l].sum + h[h[x].r].sum;
     h[x].neg_Max = h[h[x].l].neg_Max > h[h[x].r].neg_Max ? h[h[x].l].neg_Max : h[h[x].r].neg_Max;
     h[x].neg_cnt = h[h[x].l].neg_cnt + h[h[x].r].neg_cnt;
 }

 inline void Build(int &x, int l, int r)
 {
     if (!x) x = ++ tot;
     if (l == r)
     {
         h[x].sum = Abs(A[q[l]]);
         if (A[q[l]] < ) h[x].neg_Max = A[q[l]], h[x].neg_cnt = ;
             else h[x].neg_Max = -INF;
         return ;
     }
     int mid = l + r >> ;
     Build(h[x].l, l, mid);
     Build(h[x].r, mid + , r);
     update(x);
 }

 inline void apply(int x, int l, int r, LL d)
 {
     h[x].sum += d * (r - l +  - h[x].neg_cnt * );
     h[x].delta += d;
     h[x].neg_Max += d;
 }

 inline void push(int x, int l, int r)
 {
     if (h[x].delta)
     {
         int mid = l + r >> ;
         apply(h[x].l, l, mid, h[x].delta);
         apply(h[x].r, mid + , r, h[x].delta);
         h[x].delta = ;
     }
 }

 inline void Point_Modify(int x, int l, int r, LL d)
 {
     if (l == r)
     {
         h[x].sum = -h[x].sum;
         h[x].neg_Max = -INF;
         h[x].neg_cnt = ;
         return ;
     }
     int mid = l + r >> ;
     push(x, l, r);
     if (h[h[x].l].neg_Max > h[h[x].r].neg_Max)
         Point_Modify(h[x].l, l, mid, d);
     else Point_Modify(h[x].r, mid + , r, d);
     update(x);
 }

 inline void Modify(int x, int l, int r, int s, int t, LL d)
 {
     if (l == s && r == t)
     {
         while (h[x].neg_Max >= -d)
             Point_Modify(x, l, r, d);
         apply(x, l, r, d);
         return ;
     }
     push(x, l, r);
     int mid = l + r >> ;
     if (t <= mid) Modify(h[x].l, l, mid, s, t, d);
         else if (s > mid) Modify(h[x].r, mid + , r, s, t, d);
         else Modify(h[x].l, l, mid, s, mid, d), Modify(h[x].r, mid + , r, mid + , t, d);
     update(x);
 }

 inline LL Query(int x, int l, int r, int s, int t)
 {
     if (l == s && r == t) return h[x].sum;
     push(x, l, r);
     int mid = l + r >> ;
     if (t <= mid) return Query(h[x].l, l, mid, s, t);
         else if (s > mid) return Query(h[x].r, mid + , r, s, t);
         else return Query(h[x].l, l, mid, s, mid) + Query(h[x].r, mid + , r, mid + , t);
 }

 inline void Prepare()
 {
     tot = ;
     for (int i = ; i <= n; i ++)
     {
         if (Ad[i]) continue ;
         Dfn[i] = Dfn[Fa[i]] + ;
         q[] = ;
         cnt ++;
         for (int x = i; x; x = Son[x])
         {
             Ad[x] = i, Pos[x] = ++ Len[cnt];
             Num[x] = cnt, Dfn[x] = Dfn[i];
             q[++ q[]] = x;
         }
         Build(Root[cnt], , Len[cnt]);
     }
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
         freopen("4127.in", "r", stdin);
         freopen("4127.out", "w", stdout);
     #endif

     scanf("%d%d", &n, &m);
     for (int i = ; i <= n; i ++)
         scanf("%d", A + i);
     for (int i = , u, v; i < n; i ++)
     {
         scanf("%d%d", &u, &v);
         addedge(u, v);
     }
     dfs();
     Prepare();
     for (int op, u, v; m; m --)
     {
         scanf("%d", &op);
         if (op == )
         {
             LL d;
             scanf("%d%d%lld", &u, &v, &d);
             if (Dfn[u] < Dfn[v]) u = u - v, v = u + v, u = v - u;
             for (; Dfn[u] > Dfn[v]; u = Fa[Ad[u]])
                 Modify(Root[Num[u]], , Len[Num[u]], , Pos[u], d);
             for (; Ad[u] != Ad[v]; u = Fa[Ad[u]], v = Fa[Ad[v]])
             {
                 Modify(Root[Num[u]], , Len[Num[u]], , Pos[u], d);
                 Modify(Root[Num[v]], , Len[Num[v]], , Pos[v], d);
             }
             if (Pos[u] > Pos[v]) u = u - v, v = u + v, u = v - u;
             Modify(Root[Num[u]], , Len[Num[u]], Pos[u], Pos[v], d);
         }
         else
         {
             scanf("%d%d", &u, &v);
             LL ans = ;
             if (Dfn[u] < Dfn[v]) u = u - v, v = u + v, u = v - u;
             for (; Dfn[u] > Dfn[v]; u = Fa[Ad[u]])
                 ans += Query(Root[Num[u]], , Len[Num[u]], , Pos[u]);
             for (; Ad[u] != Ad[v]; u = Fa[Ad[u]], v = Fa[Ad[v]])
             {
                 ans += Query(Root[Num[u]], , Len[Num[u]], , Pos[u]);
                 ans += Query(Root[Num[v]], , Len[Num[v]], , Pos[v]);
             }
             if (Pos[u] > Pos[v]) u = u - v, v = u + v, u = v - u;
             ans += Query(Root[Num[u]], , Len[Num[u]], Pos[u], Pos[v]);
             printf("%lld\n", ans);
         }
     }

     #ifndef ONLINE_JUDGE
         fclose(stdin);
         fclose(stdout);
     #endif
     return ;
 }

4127_Gromah