SPOJ 2713 Can you answer these queries IV

SPOJ_2713

    查询区间和很容易用线段树实现，而开方操作乍一看是没法用下传lazy标记的办法实现的，但仔细想一下，实际上每个数最多被开方的次数也是很有限的，只要最后开成1或者0，那么之后都不会再变了。

    因此我们可以用一个finish标记表示当前区间内是否还有数在开方后发生变化，如果有的话finish为0，否则为1，每次开方都沿finish为0的区间更新到底，由于被开方次数有限，所以整体的复杂度还是可以承受的。

#include<stdio.h>
#include<string.h>
#include<math.h>
#define MAXD 100010
long long sum[4 * MAXD], a[MAXD];
int N, finish[4 * MAXD];
void Swap(int &x, int &y)
{
    int t;
    t = x, x = y, y = t;
}
void update(int cur)
{
    int ls = cur << 1, rs = cur << 1 | 1;
    sum[cur] = sum[ls] + sum[rs];
    if(finish[ls] && finish[rs])
        finish[cur] = 1;
    else
        finish[cur] = 0;
}
void build(int cur, int x, int y)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
    if(x == y)
    {
        sum[cur] = a[x];
        if(a[x] == 0 || a[x] == 1)
            finish[cur] = 1;
        else
            finish[cur] = 0;
        return ;
    }
    build(ls, x, mid);
    build(rs, mid + 1, y);
    update(cur);
}
void init()
{
    int i;
    for(i = 1; i <= N; i ++)
        scanf("%lld", &a[i]);
    build(1, 1, N);
}
void dfs(int cur, int x, int y)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
    if(x == y)
    {
        sum[cur] = (long long)sqrt(sum[cur] + 0.5);
        if(sum[cur] == 0 || sum[cur] == 1)
            finish[cur] = 1;
        return ;
    }
    if(!finish[ls])
        dfs(ls, x, mid);
    if(!finish[rs])
        dfs(rs, mid + 1, y);
    update(cur);
}
void refresh(int cur, int x, int y, int s, int t)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
    if(x >= s && y <= t)
    {
        if(!finish[cur])
            dfs(cur, x, y);
        return ;
    }
    if(mid >= s)
        refresh(ls, x, mid, s, t);
    if(mid + 1 <= t)
        refresh(rs, mid + 1, y, s, t);
    update(cur);
}
long long Search(int cur, int x, int y, int s, int t)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
    if(x >= s && y <= t)
        return sum[cur];
    if(mid >= t)
        return Search(ls, x, mid, s, t);
    else if(mid + 1 <= s)
        return Search(rs, mid + 1, y, s, t);
    else
        return Search(ls, x, mid, s, t) + Search(rs, mid + 1, y, s, t);
}
void solve()
{
    int i, j, k, q, x, y;
    scanf("%d", &q);
    for(i = 0; i < q; i ++)
    {
        scanf("%d%d%d", &k, &x, &y);
        if(x > y)
            Swap(x, y);
        if(k == 0)
            refresh(1, 1, N, x, y);
        else
            printf("%lld\n", Search(1, 1, N, x, y));
    }
}
int main()
{
    int t = 0;
    while(scanf("%d", &N) == 1)
    {
        init();
        printf("Case #%d:\n", ++ t);
        solve();
        printf("\n");
    }
    return 0;
}