• POJ3468---A Simple Problem with Integers


    此题简单的做法自然是 线段树 或树状数组,splay只是为了练手。。依旧 是学习bin神的模板,写了一发之后理解更深了。

      1 #include <set>
      2 #include <map>
      3 #include <cmath>
      4 #include <ctime>
      5 #include <queue>
      6 #include <stack>
      7 #include <cstdio>
      8 #include <string>
      9 #include <vector>
     10 #include <cstdlib>
     11 #include <cstring>
     12 #include <iostream>
     13 #include <algorithm>
     14 using namespace std;
     15 typedef unsigned long long ull;
     16 typedef long long ll;
     17 const int inf = 0x3f3f3f3f;
     18 const double eps = 1e-8;
     19 const int maxn = 1e5+100;
     20 ll add[maxn],sum[maxn];
     21 int ch[maxn][2],siz[maxn],key[maxn],a[maxn];
     22 int pre[maxn];
     23 int tot1,root;
     24 void new_node(int &r,int father,int k)
     25 {
     26     r = ++tot1;
     27     pre[r] = father;
     28     siz[r] = 1;
     29     add[r] = 0;
     30     key[r] = k;
     31     sum[r] = k;
     32     ch[r][0] = ch[r][1] = 0;
     33 }
     34 
     35 void update(int r,int val)
     36 {
     37     if (r == 0)
     38         return ;
     39     add[r] += val;
     40     key[r] += val;
     41     sum[r] += (ll)val * siz[r];
     42 }
     43 
     44 void push_down(int x)
     45 {
     46     if (add[x])
     47     {
     48         update(ch[x][0],add[x]);
     49         update(ch[x][1],add[x]);
     50         add[x] = 0;
     51     }
     52 }
     53 void push_up(int x)
     54 {
     55     siz[x] = siz[ch[x][0]] + siz[ch[x][1]] + 1;
     56     sum[x] = sum[ch[x][0]] + sum[ch[x][1]] + key[x];
     57 }
     58 void build(int &x,int l,int r,int father)
     59 {
     60     if (l > r)
     61         return;
     62     int mid = (l + r) >> 1;
     63     new_node(x,father,a[mid-1]);
     64     build(ch[x][0],l,mid-1,x);
     65     build(ch[x][1],mid+1,r,x);
     66     push_up(x);
     67 }
     68 void init(int n)
     69 {
     70     tot1 = 0;
     71     root = 0;
     72     new_node(root,0,-1);
     73     new_node(ch[root][1],root,-1);
     74     for (int i = 0; i < n; i++)
     75         scanf ("%d",a+i);
     76     build(ch[ch[root][1]][0],1,n,ch[root][1]);
     77     push_up(ch[root][1]);
     78     push_up(root);
     79 }
     80 void Rotate(int x,int kind)
     81 {
     82     int y = pre[x];
     83     push_down(y);
     84     push_down(x);              // 标记下传
     85 
     86     ch[y][!kind] = ch[x][kind];
     87     pre[ch[x][kind]] = y;
     88     ch[x][kind] = y;
     89     if (pre[y])
     90         ch[pre[y]][ch[pre[y]][1] == y] = x;
     91     pre[x] = pre[y];
     92     pre[y] = x;
     93     push_up(y);               //     ???
     94 }
     95 void Splay(int r,int goal)
     96 {
     97     push_down(r);
     98     while (pre[r] != goal)
     99     {
    100         if (pre[pre[r]] == goal)
    101             Rotate(r,ch[pre[r]][0] == r);
    102         else
    103         {
    104             int y = pre[r];
    105             int kind = (ch[pre[y]][0] == y);
    106             if (ch[y][kind] == r)
    107             {
    108                 Rotate(r,!kind);
    109                 Rotate(r,kind);
    110             }
    111             else
    112             {
    113                 Rotate(y,kind);
    114                 Rotate(r,kind);
    115             }
    116         }
    117     }
    118     push_up(r);
    119     if (goal == 0)
    120         root = r;
    121 }
    122 int Get_kth(int r,int k)
    123 {
    124     push_down(r);
    125     int t = siz[ch[r][0]] + 1;
    126     if (k == t)
    127         return r;
    128     if (t > k)
    129         Get_kth(ch[r][0],k);
    130     else
    131         Get_kth(ch[r][1],k-t);
    132 }
    133 void ADD(int l,int r,int val)
    134 {
    135     Splay(Get_kth(root,l),0);
    136     Splay(Get_kth(root,r+2),root);
    137     update(ch[ch[root][1]][0],val);
    138     push_up(ch[root][1]);
    139     push_up(root);
    140 }
    141 ll query(int l,int r)
    142 {
    143     Splay(Get_kth(root,l),0);
    144     Splay(Get_kth(root,r+2),root);
    145     return sum[ch[ch[root][1]][0]];
    146 }
    147 int main(void)
    148 {
    149     #ifndef ONLINE_JUDGE
    150         freopen("in.txt","r",stdin);
    151     #endif
    152     int n,q;
    153     while (~scanf ("%d%d",&n,&q))
    154     {
    155         init(n);
    156         char op[8];
    157         while (q--)
    158         {
    159             scanf ("%s",op);
    160             if (op[0] == 'C')
    161             {
    162                 int x,y,v;
    163                 scanf ("%d%d%d",&x,&y,&v);
    164                 ADD(x,y,v);
    165             }
    166             if (op[0] == 'Q')
    167             {
    168                 int x,y;
    169                 scanf ("%d%d",&x,&y);
    170                 printf("%lld
    ",query(x,y));
    171             }
    172         }
    173     }
    174     return 0;
    175 }
    View Code
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  • 原文地址:https://www.cnblogs.com/oneshot/p/4074885.html
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