PKU 2528 Mayor's posters

第三次写这个题了，这次不再用线段树，因为估计了暴力能过（复杂度的估计不太明显，最坏不超过10^8，实际很可能更低或者大部分情况下更低(O(n))）！

235ms，超时了几次，猜测是因为memset，所以加了个位优化，AC。

# include <stdio.h>
# include <stdlib.h>
# include <string.h>

# define id(x) ((x)>>3)
# define of(x) ((x)&0x7)
# define get(x) ((h[id(x)]>>of(x)) & 0x1)
# define set(x) (h[id(x)] |= (0x1 << of(x)))

# define MAXN 10005

int n, m, cols;
int l[MAXN], r[MAXN];
int p[MAXN << 2];
short c[MAXN << 2];
char h[(MAXN >> 3) + 5];

/***************************************************************/
int bins(int *v, int k)
{
    int high = m-1, low = 0, mid;
    
    while (low < high)
    {
        mid = low + ((high-low)>>1);
        if (v[mid] == k) return mid;
        else if (v[mid] > k) high = mid - 1;
        else low = mid + 1;
    }
    return low;
}

int cmp(const void *x, const void *y)
{
    return *(int*)x - *(int*)y;
}

/***************************************************************/
void solve(void)
{
    int x, y, i, j, ans;
    
    cols = 1;
    memset(c, 0, sizeof(c[0])*(m>>3));
    for (i = 0; i < n; ++i, ++cols)
    {
        x = bins(p, l[i]), y = bins(p, r[i]);
        for (j = x; j <= y; ++j) c[j] = cols;
    }
    ans = 0;
    memset(h, 0, sizeof(char)*((n+8)>>3));
    for (i = 0; i < m+5; ++i)
    {
        if (c[i] && !get(c[i])) set(c[i]), ++ans;
    }
    printf("%d\n", ans);
}

int main()
{
    int T, i, k;
    
    scanf("%d", &T);
    while (T--)
    {
        m = 0;
        scanf("%d", &n);
        for (i = 0; i < n; ++i)
        {
            scanf("%d%d", &l[i], &r[i]);
            p[m++] = l[i], p[m++] = r[i];
        }
        qsort(p, m, sizeof(p[0]), cmp);
        k = m, m = 1;
        for (i = 1; i < k; ++i)
            if (p[i] == p[m-1]) continue;
            else p[m++] = p[i];
        k = m;
        for (i = 1; i < k; ++i)
            if (p[i] != p[i-1]+1) p[m++] = p[i-1]+1;
        qsort(p, m, sizeof(p[0]), cmp);
        solve();
    }
    
    return 0;
}