HDU 1540（線段樹，以一點擴充開的最大連續區間）

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 785    Accepted Submission(s): 245

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

 /************************************************************************</p><p> 此題線段樹最經典的部分是更新部分，也是這題最難的部分，要考慮全面</p><p>線段樹結構體定義三個變數 len，l_len, r_len </p><p>分別表示這段最長的連續區間，從左節點開始的長度，從右節點開始的長度</p><p>此題一直WA。。。做的時候考慮不全面，參考了一下大牛BLOG的代碼，</p><p>發現少考慮了一些，改了才AC的~ - -|</p><p>************************************************************************/</p><p>#include <iostream><br />#include <vector><br />using namespace std;<br />#define N 50005<br />#define max(a,b) (a>b?a:b)</p><p>struct LineTree<br />{<br />int r,l;<br />int len,l_len,r_len;<br />}seg_tree[N*5];</p><p>char ch;</p><p>void Build(int root,int a,int b)<br />{<br />seg_tree[root].l=a;<br />seg_tree[root].r=b;<br />int l=(b-a+1);<br />seg_tree[root].len=l;<br />seg_tree[root].l_len=l;<br />seg_tree[root].r_len=l;<br />if(a==b) return;<br />int mid=(a+b)>>1;<br />Build(root<<1,a,mid);<br />Build((root<<1)+1,mid+1,b);<br />}</p><p>void Update(int root,int x)<br />{<br />if(seg_tree[root].l==seg_tree[root].r)<br />{<br />if(ch=='D')<br />seg_tree[root].len=seg_tree[root].l_len=seg_tree[root].r_len=0;<br />else<br />seg_tree[root].len=seg_tree[root].l_len=seg_tree[root].r_len=1;<br />return;<br />}<br />int mid=(seg_tree[root].l+seg_tree[root].r)>>1;<br />if(x<=mid) Update((root<<1),x);<br />else Update((root<<1)+1,x);<br />seg_tree[root].len=max(seg_tree[root<<1].r_len+seg_tree[(root<<1)+1].l_len,max(seg_tree[root<<1].len,seg_tree[(root<<1)+1].len));<br />seg_tree[root].l_len=seg_tree[root<<1].l_len;<br />seg_tree[root].r_len=seg_tree[(root<<1)+1].r_len;<br />if(seg_tree[root].l_len==(mid-seg_tree[root].l+1))<br />seg_tree[root].l_len+=seg_tree[(root<<1)+1].l_len;<br />if(seg_tree[root].r_len==(seg_tree[root].r-mid))<br />seg_tree[root].r_len+=seg_tree[root<<1].r_len;<br />}</p><p>int Search(int root,int x)<br />{<br />if(seg_tree[root].l==seg_tree[root].r || seg_tree[root].len==0 || seg_tree[root].len==(seg_tree[root].r-seg_tree[root].l+1))<br />{<br />return seg_tree[root].len;<br />}<br />int mid=(seg_tree[root].l+seg_tree[root].r)>>1;<br />if(x<=mid)<br />{<br />if(x > (mid-seg_tree[root<<1].r_len)) return seg_tree[root<<1].r_len+Search((root<<1)+1,mid+1);<br />else return Search(root<<1,x);<br />}<br />else<br />{<br />if(x<=(mid+seg_tree[(root<<1)+1].l_len)) return seg_tree[(root<<1)+1].l_len+Search(root<<1,mid);<br />else return Search((root<<1)+1,x);<br />}<br />}</p><p>int main()<br />{<br />int n,m,v;<br />vector<int> node;<br />while (~scanf("%d%d",&n,&m))<br />{<br />Build(1,1,n);<br />getchar();<br />node.clear();<br />while (m--)<br />{<br />scanf("%c%*c",&ch);<br />if(ch=='D')<br />{<br />scanf("%d%*c",&v);<br />node.push_back(v);<br />Update(1,v);<br />}<br />else if(ch=='R')<br />{<br />if(!node.empty())<br />{<br />v=node.back();<br />node.pop_back();<br />Update(1,v);<br />}</p><p>}<br />else<br />{<br />scanf("%d%*c",&v);<br />printf("%d/n",Search(1,v));<br />}<br />}<br />}<br />return 0;<br />}