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Solution:
根据树的遍历道的时间给树的节点编号,记录下进入节点和退出节点的时间。这个时间区间覆盖了这个节点的所有子树,可以当做连续的区间利用线段树进行操作。
/* 线段树 */ #pragma comment(linker, "/STACK:102400000,102400000") #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #define lson x<<1 #define rson x<< 1 | 1 using namespace std; const int MAX = 111111; int w[MAX], to[MAX]; int tl[MAX << 3], tr[MAX << 3], flag[MAX << 3], sum[MAX << 3][2]; int l, r; struct Edge { int v, ne; } edge[MAX << 1]; int head[MAX], cnt, num; void Dfs ( int u ) { w[u] = ++num; for ( int i = head[u]; i != 0; i = edge[i].ne ) { Dfs ( edge[i].v ); } to[u] = num; } void addE ( int u, int v ) { edge[++cnt].v = v; edge[cnt].ne = head[u], head[u] = cnt; } void build ( int x, int l, int r ) { flag[x] = sum[x][0] = sum[x][1] = 0; tl[x] = l, tr[x] = r; if ( l == r ) { sum[x][0] = 1; return ; } int mid = ( l + r ) >> 1; build ( lson, l, mid ), build ( rson, mid + 1, r ); sum[x][0] += sum[lson][0] + sum[rson][0]; } void push ( int x ) { if ( flag[x] == 0 ) return; flag[x] = 0; flag[lson] ^= 1; swap ( sum[lson][0], sum[lson][1] ); flag[rson] ^= 1; swap ( sum[rson][0], sum[rson][1] ); } void updata ( int x ) { sum[x][0] = sum[lson][0] + sum[rson][0]; sum[x][1] = sum[lson][1] + sum[rson][1]; } void modify ( int x ) { if ( tl[x] >= l && tr[x] <= r ) { flag[x] ^= 1; swap ( sum[x][0], sum[x][1] ); return ; } push ( x ); int mid = ( tl[x] + tr[x] ) >> 1; if ( l <= mid ) modify ( lson ); if ( r > mid ) modify ( rson ); updata ( x ); } int query ( int x ) { if ( tl[x] >= l && tr[x] <= r ) { return sum[x][1]; } push ( x ); int mid = ( tl[x] + tr[x] ) >> 1, ans = 0; if ( l <= mid ) ans += query ( lson ); if ( r > mid ) ans += query ( rson ); updata ( x ); return ans; } int n, m; void init() { num = cnt = 0; memset ( head, 0, sizeof head ); } int main() { while ( scanf ( "%d %d", &n, &m ) != EOF ) { init(); for ( int i = 2, v; i <= n; i++ ) { scanf ( "%d", &v ); addE ( v, i ); } Dfs ( 1 ); scanf ( "%d", &m ); build ( 1, 1, n ); char cmd[10]; for ( int i = 1, u; i <= m; ++i ) { scanf ( "%s %d", cmd, &u ); l = w[u], r = to[u]; if ( cmd[0] == ‘o‘ ) { modify ( 1 ); } else { printf ( "%d\n", query ( 1 ) ); } } putchar(10); } }
zoj 3686 A Simple Tree Problem (线段树)
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原文地址:http://www.cnblogs.com/keam37/p/4661681.html