hdu 3397 Sequence operation(线段树)

题目链接:hdu 3397 Sequence operation

题目大意：给定一个01序列，5种操作:

• 0 a b：将区间a，b上的数置为0
• 1 a b：将区间a，b上的数置为1
• 2 a b：将区间a，b上的元素0变1，1变0
• 3 a b：查询a，b中1的个数
• 4 a b：查询a，b中最大连续1的个数

解题思路：区间合并+成段更新。

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

using namespace std;

const int maxn = 100005;

int N, M, a[maxn];

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
struct Node {
    int l, r, filp, set;
    int lc[2], rc[2], sc[2];
    int cnt[2];

    int length() {
        return r - l + 1;
    }
    void maintain(int d) {
        set = d;
        filp = 0;
        for (int i = 0; i < 2; i++)
            lc[i] = rc[i] = sc[i] = cnt[i] = (i == d ? r - l + 1 : 0);
    }
    void splay() {
        if (set != -1)
            set ^= 1;
        else
            filp ^= 1;

        swap(lc[0], lc[1]);
        swap(rc[0], rc[1]);
        swap(sc[0], sc[1]);
        swap(cnt[0], cnt[1]);
    }
}nd[maxn << 2];

void pushup(int u) {
    for (int i = 0; i < 2; i++) {
        nd[u].cnt[i] = nd[lson(u)].cnt[i] + nd[rson(u)].cnt[i];
        nd[u].lc[i] = nd[lson(u)].lc[i] + (nd[lson(u)].lc[i] == nd[lson(u)].length() ? nd[rson(u)].lc[i] : 0);
        nd[u].rc[i] = nd[rson(u)].rc[i] + (nd[rson(u)].rc[i] == nd[rson(u)].length() ? nd[lson(u)].rc[i] : 0);
        nd[u].sc[i] = max(max(nd[lson(u)].sc[i], nd[rson(u)].sc[i]), nd[lson(u)].rc[i] + nd[rson(u)].lc[i]);
    }
}

void pushdown (int u) {
    if (nd[u].filp) {
        nd[lson(u)].splay();
        nd[rson(u)].splay();
        nd[u].filp = 0;
    } else if (nd[u].set != -1) {
        nd[lson(u)].maintain(nd[u].set);
        nd[rson(u)].maintain(nd[u].set);
        nd[u].set = -1;
    }
}

void build(int u, int l, int r) {
    nd[u].l = l, nd[u].r = r;
    nd[u].filp = 0, nd[u].set = -1;

    if (l == r) {
        nd[u].maintain(a[l]);
        return;
    }

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

void modify (int u, int l, int r, int v) {
    if (l <= nd[u].l && nd[u].r <= r) {
        nd[u].maintain(v);
        return;
    }

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

void modify (int u, int l, int r) {
    if (l <= nd[u].l && nd[u].r <= r) {
        nd[u].splay();
        return;
    }

    pushdown(u);
    int mid = (nd[u].l + nd[u].r) / 2;
    if (l <= mid)
        modify(lson(u), l, r);
    if (r > mid)
        modify(rson(u), l, r);
    pushup(u);
}

int query_cnt(int u, int l, int r) {
    if (l <= nd[u].l && nd[u].r <= r)
        return nd[u].cnt[1];

    pushdown(u);
    int mid = (nd[u].l + nd[u].r) / 2, ret = 0;
    if (l <= mid)
        ret += query_cnt(lson(u), l, r);
    if (r > mid)
        ret += query_cnt(rson(u), l, r);
    pushup(u);
    return ret;

}

int query_len(int u, int l, int r) {
    if (l <= nd[u].l && nd[u].r <= r)
        return nd[u].sc[1];

    pushdown(u);
    int mid = (nd[u].l + nd[u].r) / 2, ret;
    if (r <= mid)
        ret = query_len(lson(u), l, r);
    else if (l > mid)
        ret = query_len(rson(u), l, r);
    else {
        int ll = query_len(lson(u), l, r);
        int rr = query_len(rson(u), l, r);

        int A = min(nd[lson(u)].rc[1], mid - l + 1);
        int B = min(nd[rson(u)].lc[1], r - mid);
        ret = max(max(ll, rr), A + B);
    }
    pushup(u);
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d%d", &N, &M);
        for (int i = 0; i < N; i++)
            scanf("%d", &a[i]);
        build(1, 0, N - 1);

        int op, l, r;
        while (M--) {
            scanf("%d%d%d", &op, &l, &r);
            if (op == 0)
                modify(1, l, r, 0);
            else if (op == 1)
                modify(1, l, r, 1);
            else if (op == 2)
                modify(1, l, r);
            else if (op == 3)
                printf("%d\n", query_cnt(1, l, r));
            else
                printf("%d\n", query_len(1, l, r));
        }
    }
    return 0;
}