POJ 3225 Help with Intervals

POJ_3225

    对于开区间还是闭区间的问题，可以把所有的点都乘以2，这样就可以通过偶数端点判断是闭区间，通过奇数端点判断开区间。

    剩下的问题就是用线段树实现区间染色和区间取反的操作了。

#include<stdio.h>
#include<string.h>
#define MAXD 132000
#define N 131070
int tree[4 * MAXD], rev[4 * MAXD], to[4 * MAXD], a[MAXD];
void build(int cur, int x, int y)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    tree[cur] = rev[cur] = 0;
    to[cur] = -1;
    if(x == y)
        return ;
    build(ls, x, mid);
    build(rs, mid + 1, y);
}
void pushdown(int cur)
{
    int ls = cur << 1, rs = (cur << 1) | 1;
    if(to[cur] != -1)
    {
        to[ls] = to[rs] = to[cur];
        rev[ls] = rev[rs] = 0;
        tree[ls] = tree[rs] = to[cur];
        to[cur] = -1;
    }
    if(rev[cur])
    {
        rev[ls] = (rev[ls] + 1) & 1, rev[rs] = (rev[rs] + 1) & 1;
        tree[ls] ^= 1, tree[rs] ^= 1;
        rev[cur] = 0;
    }
}
void color(int cur, int x, int y, int s, int t, int c)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    if(x >= s && y <= t)
    {
        tree[cur] = to[cur] = c;
        rev[cur] = 0;
        return ;
    }
    pushdown(cur);
    if(mid >= s)
        color(ls, x, mid, s, t, c);
    if(mid + 1 <= t)
        color(rs, mid + 1, y, s, t, c);
}
void Reverse(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)
    {
        tree[cur] ^= 1;
        rev[cur] = (rev[cur] + 1) & 1;
        return ;
    }
    pushdown(cur);
    if(mid >= s)
        Reverse(ls, x, mid, s, t);
    if(mid + 1 <= t)
        Reverse(rs, mid + 1, y, s, t);
}
void dfs(int cur, int x, int y)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    if(x == y)
    {
        a[x] = tree[cur];
        return ;
    }
    pushdown(cur);
    dfs(ls, x, mid);
    dfs(rs, mid + 1, y);
}
void printresult()
{
    int i, x, y, flag = 0;
    dfs(1, 0, N);
    for(i = 0; i <= N; i ++)
    {
        if(a[i])
        {
            if(flag)
                printf(" ");
            else
                flag = 1;
            if(i & 1)
                printf("(%d,", i >> 1);
            else
                printf("[%d,", i >> 1);
            for(; a[i + 1]; i ++);
            if(i & 1)
                printf("%d)", (i + 1) >> 1);
            else
                printf("%d]", i >> 1);
        }
    }
    if(flag == 0)
        printf("empty set\n");
    printf("\n");
}
void solve()
{
    int x, y;
    char op[5], b[20];
    while(scanf("%s%s", op, b) == 2)
    {
        sscanf(b + 1, "%d,%d", &x, &y);
        if(b[0] == '(')
            x = (x << 1) | 1;
        else
            x <<= 1;
        if(b[strlen(b) - 1] == ')')
            y = (y << 1) - 1;
        else
            y <<= 1;
        if(op[0] == 'U')
            color(1, 0, N, x, y, 1);
        else if(op[0] == 'D')
            color(1, 0, N, x, y, 0);
        else if(op[0] == 'S')
            Reverse(1, 0, N, x, y);
        else if(op[0] == 'C')
        {
            if(x > 0)
                color(1, 0, N, 0, x - 1, 0);
            if(y < N)
                color(1, 0, N, y + 1, N, 0);
            Reverse(1, 0, N, x, y);
        }
        else
        {
            if(x > 0)
                color(1, 0, N, 0, x - 1, 0);
            if(y < N)
                color(1, 0, N, y + 1, N, 0);
        }
    }
    printresult();
}
int main()
{
    build(1, 0, N);
    solve();
    return 0;
}