hdu1540 && POJ2892 Tunnel Warfare，

hdu1540 && POJ2892 Tunnel Warfare，
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Tunnel WarfareTime Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4513    Accepted Submission(s): 1725


Problem DescriptionDuring 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!
 
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 9D 3D 6D 5Q 4Q 5RQ 4RQ 4

 
Sample Output

1024

 
SourcePOJ Monthly 
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這道題也是典型的線段樹區間合并的題, 主要是要記錄一段區間左邊連續的長度, 右邊連續長度, 單點更新,查詢時只要判斷要查的點是不是落在中間連續處,否則就繼續遞迴查詢

/*************************************************************************    > File Name: hdu1540.cpp    > Author: ALex    > Mail: 405045132@qq.com     > Created Time: 2015年01月11日 星期日 12時24分34秒 ************************************************************************/#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 500010;stack <int> re;struct node{int l, r;int l_sum;int r_sum;}tree[N << 2];void pushup (int p){tree[p].l_sum = tree[p << 1].l_sum;tree[p].r_sum = tree[p << 1 | 1].r_sum;if (tree[p << 1].l_sum == tree[p << 1].r - tree[p << 1].l + 1){tree[p].l_sum += tree[p << 1 | 1].l_sum;}if (tree[p << 1 | 1].r_sum == tree[p << 1 | 1].r - tree[p << 1 | 1].l + 1){tree[p].r_sum += tree[p << 1].r_sum;}}void build (int p, int l, int r){tree[p].l = l;tree[p].r = r;tree[p].l_sum = tree[p].r_sum = (r - l + 1);if (l == r){return;}int mid = (l + r) >> 1;build (p << 1, l, mid);build (p << 1 | 1, mid + 1, r);}void update (int p, int pos, int sta){if (tree[p].l == tree[p].r){tree[p].l_sum = tree[p].r_sum = sta;return;}int mid = (tree[p].l + tree[p].r) >> 1;if (pos <= mid){update (p << 1, pos, sta);}else{update (p << 1 | 1, pos, sta);}pushup (p);}int query (int p, int pos){if (tree[p].l == tree[p].r){if (tree[p].l_sum == 1){return 1;}return 0;}int mid = (tree[p].l + tree[p].r) >> 1;if (pos <= mid){if (pos >= mid - tree[p << 1].r_sum + 1){return tree[p << 1].r_sum + tree[p << 1 | 1].l_sum;}return query (p << 1, pos);}else{if (pos <= mid + tree[p << 1 | 1].l_sum){return tree[p << 1 | 1].l_sum + tree[p << 1].r_sum;}return query (p << 1 | 1, pos);}}int main(){int n, m;int x;char op[10];while (~scanf("%d%d", &n, &m)){build (1, 1, n);while (!re.empty()){re.pop();}while (m--){scanf("%s", op);if(op[0] == 'R'){if (re.empty()){continue;}update (1, re.top(), 1);re.pop();}else{scanf("%d", &x);if (op[0] == 'D'){update (1, x, 0);re.push(x);}else{printf("%d\n", query (1, x));}}}}return 0;}