題目鏈接
題意：給定一個01序列，有5種操作
0 a b [a.b]區間置為0
1 a b [a,b]區間置為1
2 a b [a,b]區間0變成1，1變成0
3 a b 查詢[a,b]區間1的個數
4 a b 查詢[a,b]區間連續1最長的長度
思路：線段樹線段合並，需要兩個延遲標記一個置為01，一個翻轉，然後由於4操作，需要記錄左邊最長0、1，右邊最長0、1，區間最長0、1，然後區間合並去搞即可
代碼：
#include#include #include using namespace std; #define lson(x) ((x<<1)+1) #define rson(x) ((x<<1)+2) const int N = 100005; int t, n, m; struct Node { int l, r, sum[3][2], num; int setv, flip; Node() { setv = -1; flip = 0; } int size() {return r - l + 1;} void gao1(int v) { setv = v; num = v * (r - l + 1); for (int i = 0; i < 3; i++) { sum[i][v] = r - l + 1; sum[i][!v] = 0; } flip = 0; } void gao2() { flip ^= 1; for (int i = 0; i < 3; i++) swap(sum[i][0], sum[i][1]); num = r - l + 1 - num; } } node[N * 4]; Node merge(Node lson, Node rson) { Node x; x.l = lson.l; x.r = rson.r; for (int i = 0; i < 2; i++) { x.sum[0][i] = lson.sum[0][i]; x.sum[1][i] = rson.sum[1][i]; x.sum[2][i] = max(lson.sum[2][i], rson.sum[2][i]); if (lson.sum[0][i] == lson.size()) x.sum[0][i] += rson.sum[0][i]; if (rson.sum[1][i] == rson.size()) x.sum[1][i] += lson.sum[1][i]; x.sum[2][i] = max(x.sum[2][i], lson.sum[1][i] + rson.sum[0][i]); } x.num = lson.num + rson.num; return x; } void pushup(int x) { node[x] = merge(node[lson(x)], node[rson(x)]); } void pushdown(int x) { if (node[x].setv != -1) { node[lson(x)].gao1(node[x].setv); node[rson(x)].gao1(node[x].setv); node[x].setv = -1; } if (node[x].flip) { node[lson(x)].gao2(); node[rson(x)].gao2(); node[x].flip = 0; } } void build(int l, int r, int x = 0) { node[x].l = l; node[x].r = r; node[x].setv = -1; node[x].flip = 0; if (l == r) { int tmp; scanf("%d", &tmp); for (int i = 0; i < 3; i++) { node[x].sum[i][tmp] = 1; node[x].sum[i][!tmp] = 0; } node[x].num = tmp; return; } int mid = (l + r) / 2; build(l, mid, lson(x)); build(mid + 1, r, rson(x)); pushup(x); } void add(int l, int r, int v, int x = 0) { if (node[x].l >= l && node[x].r <= r) { if (v == 2) node[x].gao2(); else node[x].gao1(v); return; } int mid = (node[x].l + node[x].r) / 2; pushdown(x); if (l <= mid) add(l, r, v, lson(x)); if (r > mid) add(l, r, v, rson(x)); pushup(x); } Node query(int l, int r, int x = 0) { if (node[x].l >= l && node[x].r <= r) return node[x]; int mid = (node[x].l + node[x].r) / 2; pushdown(x); Node ans; if (l <= mid && r > mid) ans = merge(query(l, r, lson(x)), query(l, r, rson(x))); else if (l <= mid) ans = query(l, r, lson(x)); else if (r > mid) ans = query(l, r, rson(x)); pushup(x); return ans; } int main() { scanf("%d", &t); while (t--) { scanf("%d%d", &n, &m); build(0, n - 1); int op, a, b; while (m--) { scanf("%d%d%d", &op, &a, &b); if (op <= 2) add(a, b, op); else if (op == 3) printf("%d\n", query(a, b).num); else if (op == 4) printf("%d\n", query(a, b).sum[2][1]); } } return 0; }
