poj 3225 Help with Intervals(线段树)

题目链接：poj 3225 Help with Intervals

题目大意：模拟集合操作，输出最终的集合。

解题思路：线段树。

• U l r：[l,r]区间置为1
• I l r：[0,l),(r,maxn]置为0
• D l r：[l,r]区间置为0
• C l r：[0,l),(r,maxn]置为0,[l,r]区间取逆
• S l r：[l，r]区间取逆。

然后基本水水的线段树，注意一下区间开和闭。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 65535 * 2;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)+1)

int lc[maxn * 4], rc[maxn * 4], set[maxn * 4], filp[maxn * 4];

inline void splay(int u) {
    filp[u] ^= 1;

    if (filp[u] && set[u] != -1) {
        filp[u] = 0;
        set[u] ^= 1;
    }
}

inline void maintain(int u, int v) {
    set[u] = v;
    filp[u] = 0;
}

inline void pushdown (int u) {
    if (set[u] != -1) {
        maintain(lson(u), set[u]);
        maintain(rson(u), set[u]);
        set[u] = -1;
    }

    if (filp[u]) {
        splay(lson(u));
        splay(rson(u));
        filp[u] = 0;
    }
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    maintain(u, -1);

    if (l == r) {
        maintain(u, 0);
        return;
    }

    int mid = (l + r) / 2;
    build (lson(u), l, mid);
    build (rson(u), mid + 1, r);
}

void modify (int u, int l, int r, int v) {
    if (l > r) return;

    if (l <= lc[u] && rc[u] <= r) {
        maintain(u, v);
        return;
    }

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        modify(lson(u), l, r, v);
    if (r > mid)
        modify(rson(u), l, r, v);
}

void change (int u, int l, int r) {
    if (l > r) return;

    if (l <= lc[u] && rc[u] <= r) {
        splay(u);
        return;
    }

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        change(lson(u), l, r);
    if (r > mid)
        change(rson(u), l, r);
}

int query (int u, int x) {
    if (lc[u] == x && x == rc[u])
        return set[u];

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (x <= mid)
        return query(lson(u), x);
    else
        return query(rson(u), x);
}

int L, R;
char op, LP, RP;

inline void put (int left, int right) {
    if (left&1)
        printf("(%d,", left/2);
    else
        printf("[%d,", left/2);

    if (right&1)
        printf("%d)", (right + 1) / 2);
    else
        printf("%d]", right / 2);
}

int main () {
    build (1, 0, maxn);

    while (~scanf("%c%*c%c%d,%d%c%*c", &op, &LP, &L, &R, &RP)) {
        L *= 2;
        R *= 2;
        if (LP == ‘(‘)
            L++;
        if (RP == ‘)‘)
            R--;

        if (op == ‘U‘) {
            modify(1, L, R, 1);
        } else if (op == ‘I‘) {
            modify(1, 0, L - 1, 0);
            modify(1, R + 1, maxn, 0);
        } else if (op == ‘D‘) {
            modify(1, L, R, 0);
        } else if (op == ‘C‘) {
            change(1, L, R);
            modify(1, 0, L - 1, 0);
            modify(1, R + 1, maxn, 0);
        } else if (op == ‘S‘)
            change(1, L, R);
    }

    bool flag = false;
    int c = 0, t;

    for (int i = 0; i <= maxn; i++) {
        int s = query(1, i);
        if (s && !flag) {
            t = i;
            flag = true;
        } else if (!s && flag) {
            if (c++)
                printf(" ");
            put(t, i-1);
            flag = false;
        }
    }

    if (c == 0)
        printf("empty set\n");
    else
        printf("\n");

    return 0;
}