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  • HDOJ 3397 Sequence operation

    难在异或对询问区间最大连续1长度的影响,每次异或时,把异或标记更新到底(cover标记不为-1),直接更新当前最大连续1区间的长度。
      1 # include <stdio.h>
      2 
      3 # define N 100005
      4 
      5 # define ls ((r)<<1)
      6 # define rs ((r)<<1|1)
      7 # define mid (((x)+(y))>>1)
      8 
      9 int n, m, a[N];
     10 int cover[N<<2];
     11 int sum[N<<2], lcon[N<<2], rcon[N<<2], tcon[N<<2];
     12 
     13 int Max(int x, int y)
     14 {
     15     return x > y ? x : y;
     16 }
     17 
     18 int Min(int x, int y)
     19 {
     20     return x < y ? x : y;
     21 }
     22 
     23 void update(int r, int x, int y)
     24 {
     25     sum[r] = sum[ls] + sum[rs];
     26     lcon[r] = lcon[ls], rcon[r] = rcon[rs];
     27     if (lcon[ls] == mid-x+1) lcon[r] += lcon[rs];
     28     if (rcon[rs] == y-mid) rcon[r] += rcon[ls];
     29     tcon[r] = Max(rcon[ls]+lcon[rs], Max(tcon[ls], tcon[rs]));
     30 }
     31 
     32 void pushdown(int r, int x, int y)
     33 {
     34     if (cover[r] != -1)
     35     {
     36         cover[ls] = cover[rs] = cover[r];
     37         sum[ls] = (cover[r] ? mid-x+1:0);
     38         sum[rs] = (cover[r] ? y-mid:0);
     39         lcon[ls] = rcon[ls] = tcon[ls] = sum[ls];
     40         lcon[rs] = rcon[rs] = tcon[rs] = sum[rs];
     41         cover[r] = -1;
     42     }
     43 }
     44 
     45 void build(int r, int x, int y)
     46 {
     47     cover[r] = -1;
     48     if (x == y)
     49     {
     50         cover[r] = sum[r] = a[x];
     51         lcon[r] = rcon[r] = tcon[r] = a[x];
     52         return ;
     53     }
     54     build(ls, x, mid);
     55     build(rs, mid+1, y);
     56     update(r, x, y);
     57 }
     58 
     59 void change(int r, int x, int y, int s, int t, int val)
     60 {
     61     if (s<=x && y<=t)
     62     {
     63         cover[r] = val;
     64         sum[r] = (val ? y-x+1:0);
     65         lcon[r] = rcon[r] = tcon[r] = sum[r];
     66         return ;
     67     }
     68     pushdown(r, x, y);
     69     if (s <= mid) change(ls, x, mid, s, t, val);
     70     if (mid+1<=t) change(rs, mid+1, y, s, t, val);
     71     update(r, x, y);
     72 }
     73 
     74 void query_sum(int r, int x, int y, int s, int t, int *ans)
     75 {
     76     if (s<=x && y<=t)
     77     {
     78         *ans += sum[r];
     79         return ;
     80     }
     81     pushdown(r, x, y);
     82     if (s <= mid) query_sum(ls, x, mid, s, t, ans);
     83     if (mid+1<=t) query_sum(rs, mid+1, y, s, t, ans);
     84 }
     85 
     86 /******************************************************/
     87 void procxor(int r, int x, int y, int s, int t)
     88 {
     89     if (s<=x && y<=t)
     90     {
     91         if (cover[r] != -1)
     92         {
     93             cover[r] ^= 1;
     94             sum[r] = (cover[r] ? y-x+1:0);
     95             lcon[r] = rcon[r] = tcon[r] = sum[r];
     96             return ;
     97         }
     98     }
     99     pushdown(r, x, y);
    100     if (s <= mid) procxor(ls, x, mid, s, t);
    101     if (mid+1 <= t) procxor(rs, mid+1, y, s, t);
    102     update(r, x, y);
    103 }
    104 
    105 int query_len(int r, int x, int y, int s, int t)
    106 {
    107     int a, b, c;
    108     if (s==x && y==t) return tcon[r];
    109     pushdown(r, x, y);
    110     if (t <= mid) return query_len(ls, x, mid, s, t);
    111     else if (s >= mid+1) return query_len(rs, mid+1, y, s, t);
    112     else
    113     {
    114         a = query_len(ls, x, mid, s, mid);
    115         b = query_len(rs, mid+1, y, mid+1, t);
    116         c = Min(rcon[ls], mid-s+1)+Min(lcon[rs], t-mid);
    117         return Max(a, Max(b, c));
    118     }
    119 }
    120 /******************************************************/
    121 void solve(void)
    122 {
    123     int i, op, s, t, ans;
    124     scanf("%d%d", &n, &m);
    125     for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
    126     build(1, 1, n);
    127     for (i = 1; i <= m; ++i)
    128     {
    129         scanf("%d%d%d", &op, &s, &t), ++s, ++t;
    130         switch(op)
    131         {
    132             case 0: change(1, 1, n, s, t, 0); break;
    133             case 1: change(1, 1, n, s, t, 1); break;
    134             case 2: procxor(1, 1, n, s, t); break;
    135             case 3: ans = 0, query_sum(1, 1, n, s, t, &ans), printf("%d\n", ans); break;
    136             case 4: ans = query_len(1, 1, n, s, t), printf("%d\n", ans); break;
    137         }
    138     }
    139 }
    140 
    141 int main()
    142 {
    143     int T;
    144     scanf("%d", &T);
    145     while (T--)    solve();
    146     return 0;
    147 }

      

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  • 原文地址:https://www.cnblogs.com/JMDWQ/p/2654532.html
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