During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones.
Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!
Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!
InputThe first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.
There are three different events described in different format shown below:
D x: The x-th village was destroyed.
Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.
R: The village destroyed last was rebuilt.
OutputOutput the answer to each of the Army commanders’ request in order on a separate line.
Sample Input
7 9 D 3 D 6 D 5 Q 4 Q 5 R Q 4 R Q 4
Sample Output
1 0 2 4
题解：每个区间有三个特征值：①以左端点为起始位置的最长连续段(全1)，②以右端点为起始位置的最长连续段(全1)，③该区间最长的连续段(全1)；
更新操作好做，注意一下当Tree[root].ls=(Tree[lson].r-Tree[lson].l+1)的时候表明，Tree[rson].ls也可以加在Tree[root].ls上。
重点是查询，个人觉得不太好理解。
1 #pragma warning(disable:4996) 2 #include<cstdio> 3 #include<stack> 4 #include<cmath> 5 #include<string> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std; 10 11 #define lson root<<1 12 #define rson root<<1|1 13 #define ll long long 14 15 const int maxn = 200005; 16 17 int n, q; 18 19 struct node { 20 int l, r; 21 int ls, rs, Max; 22 }Tree[maxn]; 23 24 void Pushup(int root) { 25 Tree[root].ls = Tree[lson].ls; 26 Tree[root].rs = Tree[rson].rs; 27 Tree[root].Max = max((Tree[lson].rs + Tree[rson].ls), max(Tree[lson].Max, Tree[rson].Max)); 28 if (Tree[lson].ls == (Tree[lson].r - Tree[lson].l + 1)) Tree[root].ls += Tree[rson].ls; 29 if (Tree[rson].rs == (Tree[rson].r - Tree[rson].l + 1)) Tree[root].rs += Tree[lson].rs; 30 } 31 32 void Build(int l, int r, int root) { 33 Tree[root].l = l; 34 Tree[root].r = r; 35 if (l == r) { 36 Tree[root].Max = Tree[root].ls = Tree[root].rs = 1; 37 return; 38 } 39 int mid = (l + r) >> 1; 40 Build(l, mid, lson); 41 Build(mid + 1, r, rson); 42 Pushup(root); 43 } 44 45 void Update(int k, int l, int r, int root, int x) { 46 if (l == r) { 47 Tree[root].Max = Tree[root].ls = Tree[root].rs = x; 48 return; 49 } 50 int mid = (l + r) >> 1; 51 if (k <= mid) Update(k, l, mid, lson, x); 52 else Update(k, mid + 1, r, rson, x); 53 Pushup(root); 54 } 55 56 int Query(int k, int l, int r, int root) { 57 if (l == r || Tree[root].Max == 0 || Tree[root].Max == (r - l + 1)) return Tree[root].Max; 58 int mid = (l + r) >> 1; 59 if (k <= mid) { 60 if (Tree[lson].rs >= (mid - k + 1)) return Tree[lson].rs + Tree[rson].ls; 61 else return Query(k, l, mid, lson); 62 } 63 else { 64 if (Tree[rson].ls >= (k - mid)) return Tree[rson].ls + Tree[lson].rs; 65 else return Query(k, mid + 1, r, rson); 66 } 67 } 68 69 int main() 70 { 71 while (scanf("%d%d", &n, &q) != EOF) { 72 Build(1, n, 1); 73 74 stack<int> s; 75 76 char op[2]; 77 for (int i = 1; i <= q; i++) { 78 scanf("%s", op); 79 if (op[0] == 'D') { 80 int x; 81 scanf("%d", &x); 82 s.push(x); 83 Update(x, 1, n, 1, 0); 84 } 85 else if (op[0] == 'R') { 86 if (s.empty()) continue; 87 int x = s.top(); 88 s.pop(); 89 Update(x, 1, n, 1, 1); 90 } 91 else { 92 int x; 93 scanf("%d", &x); 94 printf("%d ", Query(x, 1, n, 1)); 95 } 96 } 97 } 98 return 0; 99 }