HDU 3911 Black and White (line segment tree)

Connection: HDU 3911 black and white

A sequence is given and M operations are performed;

• 0 l r: indicates the maximum number of consecutive 1 s in query L and R.
• 1 l r: returns the inverse of the number in the range of L and R.

Solution: A Question type of the Line Segment tree. The interval is merged. Because there is an inverse operation, we need to maintain six values for each node, including the length of the longest continuous sequence, the longest continuous length on the left and right. It should be noted that if the maximum value of the queried range is from R [lson] + L [rson], you must determine whether the ratio length is greater than R-l + 1. I didn't pay attention to it at the beginning, so Wa went online and searched for other people's questions. I found that many people did not pushup in the query to update the current node information. This is definitely not a problem, but Pushdown exists, you must update the current node information.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e5 + 5;int N, M, a[maxn];#define lson(x) ((x)<<1)#define rson(x) (((x)<<1)|1)int lc[maxn << 2], rc[maxn << 2], filp[maxn << 2];int L[maxn << 2][2], R[maxn << 2][2], S[maxn << 2][2];void maintain (int u) {    filp[u] ^= 1;    swap(L[u][0], L[u][1]);    swap(R[u][0], R[u][1]);    swap(S[u][0], S[u][1]);}void pushup(int u) {    for (int i = 0; i < 2; i++) {        S[u][i] = max(max(S[lson(u)][i], S[rson(u)][i]), R[lson(u)][i] + L[rson(u)][i]);        L[u][i] = L[lson(u)][i] + (L[lson(u)][i] == rc[lson(u)] - lc[lson(u)] + 1 ? L[rson(u)][i] : 0);        R[u][i] = R[rson(u)][i] + (R[rson(u)][i] == rc[rson(u)] - lc[rson(u)] + 1 ? R[lson(u)][i] : 0);    }}void pushdown (int u) {    if (filp[u]) {        maintain(lson(u));        maintain(rson(u));        filp[u] = 0;    }}void build (int u, int l, int r) {    lc[u] = l;    rc[u] = r;    filp[u] = 0;    if (l == r) {        int d = a[l];        L[u][d] = R[u][d] = S[u][d] = 1;        L[u][d^1] = R[u][d^1] = S[u][d^1] = 0;        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) {    if (l <= lc[u] && rc[u] <= r) {        maintain(u);        return;    }    pushdown(u);    int mid = (lc[u] + rc[u]) / 2;    if (l <= mid)        modify(lson(u), l, r);    if (r > mid)        modify(rson(u), l, r);    pushup(u);}int query (int u, int l, int r) {    if (l <= lc[u] && rc[u] <= r)        return S[u][1];    pushdown(u);    int mid = (lc[u] + rc[u]) / 2, ret;    if (r <= mid)        ret = query(lson(u), l, r);    else if (l > mid)        ret = query(rson(u), l, r);    else {        int ll = query(lson(u), l, r);        int rr = query(rson(u), l, r);        int a = min(L[rson(u)][1], r - mid);        int b = min(R[lson(u)][1], mid - l + 1);        ret = max( max(ll, rr), a + b);    }    pushup(u);    return ret;}int main () {    int x, l, r;    while (scanf("%d", &N) == 1) {        for (int i = 1; i <= N; i++)            scanf("%d", &a[i]);        build (1, 1, N);        scanf("%d", &M);        while (M--) {            scanf("%d%d%d", &x, &l, &r);            if (x)                modify(1, l, r);            else                printf("%d\n", query(1, l, r));        }    }    return 0;}

HDU 3911 Black and White (line segment tree)