题意:有一个长板子,多次操作,有两种操作,第一种是给从a到b那段染一种颜色c,另一种是询问a到b有多少种不同的颜色。
这题更加让我理解线段树的结构了,特别是lazy思想的运用。
事实上lazy思想就是个懒人的标记,若对于这个结点lazy标记为true,就代表不需要继续查找缩小的区间了。
主要是在更新结点的地方,若填充整个区间时,标记lazy,则在下次其他的更新操作时,若结点为ture,则在更新操作中。
为了控制标记,取消原来结点的标记false,表示此节点不可用,即该结点代表的线段中有不同的取值,然后在左右子树中标记lazy,直到填充整个区间。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> using namespace std; #define maxn 100004 struct Node { int color; int l, r; Node *pleft, *pright; bool end; } tree[maxn * 3]; int n, t, o, ncount; void buildtree(Node *proot, int l, int r) { proot->l = l; proot->r = r; proot->color = 1; proot->end = true; if (l == r) return; int mid = (l + r) / 2; ncount++; proot->pleft = tree + ncount; ncount++; proot->pright = tree + ncount; buildtree(proot->pleft, l, mid); buildtree(proot->pright, mid + 1, r); } void paint(Node *proot, int l, int r, int color) { if (proot->l == l && proot->r == r) { proot->end = true; proot->color = color; return; } if (proot->end) //lazy思想,当一次取整块lr区间时,标记end, { //然后下次操作遇到lr区间但不是目标区间时,消去该节点的end标记 proot->end = false; //为了简化后面的操作,直接给左右子树进行相应操作并同意标记end proot->pleft->color = proot->color; proot->pleft->end = true; proot->pright->color = proot->color; proot->pright->end = true; } int mid = (proot->l + proot->r) / 2; if (r <= mid) paint(proot->pleft, l, r, color); else if(l > mid) paint(proot->pright, l, r, color); else { paint(proot->pleft, l, mid, color); paint(proot->pright, mid +1, r, color); } proot->color = proot->pleft->color | proot->pright->color; //位运算 } int query(Node *proot, int l, int r) { if (proot->end) //若父节点已经被标记.便不需要往下找 return proot->color; if (proot->l == l && proot->r == r) return proot->color; int mid = (proot->l + proot->r) / 2; if (r <= mid) return query(proot->pleft, l, r); else if(l > mid) return query(proot->pright, l, r); return query(proot->pleft, l, mid) | query(proot->pright, mid + 1, r); } int countbit(int a) { int x = 1; int ret = 0; for (int i = 0; i < 32; i++, x <<= 1) if (x & a) ret++; return ret; } int main() { // freopen("t.txt", "r", stdin); ncount = 0; scanf("%d%d%d", &n, &t, &o); getchar(); buildtree(tree, 1, n); for (int i = 0; i < o; i++) { char order; int l, r, c; scanf("%c", &order); if (order == 'C') { scanf("%d%d%d", &l, &r, &c); if (l > r) swap(l, r); paint(tree, l, r, 1 << (c - 1)); } else { scanf("%d%d", &l, &r); if (l > r) swap(l, r); printf("%d\n", countbit(query(tree, l, r))); } getchar(); } return 0; }