11月2日模拟赛题解

T1 黑白图像压缩

简单模拟题 略

T2 校门外的区间

题意

解法：

问题可以转化为线段树区间赋\(0/1\)值，区间取反操作

对于开闭区间的处理，可以转化为将每两个点中间加一个点，表示“这两个点之间的区间”的问题。

注意对于懒标记的\(pushdown\)的处理问题

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 70010
char ch[20], op[10];
int pos;
int add[maxn * 4], tag[maxn * 4];
void pushdown(int k, int l, int r)
{
    if(tag[k] == 1)
    {
        if(add[k << 1] == -1) tag[k << 1] ^= 1;
        else add[k << 1] ^= 1;
        if(add[k << 1 | 1] == -1) tag[k << 1 | 1] ^= 1;
        else add[k << 1 | 1] ^= 1;
        tag[k] = 0;
    }
    if(add[k] == -1) return;
    add[k << 1] = add[k];
    add[k << 1 | 1] = add[k];
    add[k] = -1;
    return;
}
void modify(int k, int l, int r, int x, int y, int v)
{
    if(x <= l && r <= y)
    {
        add[k] = v;
        return;
    }
    pushdown(k, l, r);
    int mid = (l + r) >> 1;
    if(mid >= x) modify(k << 1, l, mid, x, y, v);
    if(mid < y) modify(k << 1 | 1, mid + 1, r, x, y, v);
    return;
}
void update(int k, int l, int r, int x, int y)
{
    if(x <= l && r <= y)
    {
        if(add[k] == -1) tag[k] ^= 1;
        else add[k] ^= 1;
        return;
    }
    pushdown(k, l, r);
    int mid = (l + r) >> 1;
    if(mid >= x) update(k << 1, l, mid, x, y);
    if(mid < y) update(k << 1 | 1, mid + 1, r, x, y);
    return;
}
int query(int k, int l, int r, int x)
{
    if(l == r)
    {
        if(add[k] == -1) return tag[k];
        else return add[k];
    }
    pushdown(k, l, r);
    int mid = (l + r) >> 1;
    if(mid >= x) return query(k << 1, l, mid, x);
    if(mid < x) return query(k << 1 | 1, mid + 1, r, x);
}
int main()
{
    memset(add, -1, sizeof(add));
    int l, r;
    int n = 65535;
    while(scanf("%s", op + 1) != EOF)
    {
        scanf("%s", ch + 1);
        int len = strlen(ch + 1);
        for(int i = 1; i <= len; i++) if(ch[i] == ',') pos = i;
        l = r = 0;
        for(int i = 2; i < pos; i++) l = l * 10 + ch[i] - '0';
        for(int i = pos + 1; i < len; i++) r = r * 10 + ch[i] - '0';
        if(ch[1] == '[') l = l * 2 + 1; else l = l * 2 + 2;
        if(ch[len] == ')') r = r * 2; else r = r * 2 + 1;
        if(op[1] == 'U') modify(1, 1, 2 * (n + 1), l, r, 1);
        if(op[1] == 'I')
        {
            if(l != 1) modify(1, 1, 2 * n + 2, 1, l - 1, 0);
            if(r != 2 * n + 2) modify(1, 1, 2 * n + 2, r + 1, 2 * n + 2, 0);
        }
        if(op[1] == 'D') modify(1, 1, 2 * n + 2, l, r, 0);
        if(op[1] == 'C')
        {
            update(1, 1, 2 * n + 2, 1, 2 * n + 2);
            if(l != 1) modify(1, 1, 2 * n + 2, 1, l - 1, 0);
            if(r != 2 * n + 2) modify(1, 1, 2 * n + 2, r + 1, 2 * n + 2, 0);
        }
        if(op[1] == 'S')
        {
            update(1, 1, 2 * n + 2, l, r);
        }
    }
    int lst = 1;
    int pd = 0; int now = 0;
    while(query(1, 1, 2 * n + 2, lst) == 0 && lst <= 2 * n + 1) lst++;
    now = lst;
    while(now <= 2 * n + 1)
    {
        int x = query(1, 1, n * 2 + 2, now + 1);
        if(x == 0)
        {
            if(lst % 2 == 1)
            {
                printf("[%d,", lst / 2);
            } else printf("(%d,", (lst - 1) / 2);
            if(now % 2 == 1)
            {
                printf("%d]", now / 2);
            } else printf("%d)", now / 2);
            pd = 1; printf(" ");
            lst = now + 1;
            while(query(1, 1, 2 * n + 2, lst) == 0 && lst <= (n * 2 + 1)) lst++;
            now = lst + 1;
        }
        else
        {
            now++;
            continue;
        }
    }
    if(!pd) printf("empty set\n");
    return 0;
}