題目鏈接

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Description

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!

Input

The 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.

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Output

Output 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

Hint

An illustration of the sample input:

OOOOOOO
D 3 OOXOOOO
D 6 OOXOOXO
D 5 OOXOXXO
R OOXOOXO
R OOXOOOO

Source

POJ Monthly--2006.07.30, updog 題意：有n個村莊編號為1，2，3...n 它們按照序號一一相連，現(xiàn)在有m次操作，有以下幾種操作： 1、D ?x ?表示將x號村莊摧毀。 2、Q ?x ?表示查詢x村莊能到達的村莊數(shù)（包括x村莊）。 3、R ? ? ?表示修復(fù)最近一個被摧毀的村莊。 每次查詢輸出一個值。 思路：線段樹單點更新、區(qū)間合并，用棧存儲被摧毀的村莊號。 代碼如下： #include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <stack> using namespace std; const int maxn=50005; stack<int> s; struct Node{int l,r,m; }tr[4*maxn];void build(int i,int l,int r) {tr[i].l=tr[i].r=tr[i].m=r-l+1;if(l==r) return;int mid=(l+r)/2;build(2*i,l,mid);build(2*i+1,mid+1,r); }void update(int i,int l,int r,int x,int y) {if(l==r){tr[i].l=tr[i].r=tr[i].m=y;return;}int mid=(l+r)/2;if(x<=mid) update(2*i,l,mid,x,y);else update(2*i+1,mid+1,r,x,y);if(tr[2*i].m==mid-l+1) tr[i].l=tr[2*i].m+tr[2*i+1].l;else tr[i].l=tr[2*i].l;if(tr[2*i+1].m==r-mid) tr[i].r=tr[2*i+1].m+tr[2*i].r;else tr[i].r=tr[2*i+1].r;tr[i].m=max(max(tr[2*i].m,tr[2*i+1].m),tr[2*i].r+tr[2*i+1].l); }int query(int i,int l,int r,int x) {int sum=0;if(l==r) return tr[i].m;if(r-l+1==tr[i].m) return tr[i].m;int mid=(l+r)/2;if(x<=mid){if(mid-tr[2*i].r+1<=x)return tr[2*i].r+tr[2*i+1].l;else return query(2*i,l,mid,x);}else {if(tr[2*i+1].l+mid>=x)return tr[2*i].r+tr[2*i+1].l;else return query(2*i+1,mid+1,r,x);} }int main() {int n,m;while(scanf("%d",&n)!=EOF){scanf("%d",&m);build(1,1,n);int x;char str[5];while(!s.empty()) s.pop();while(m--){scanf("%s",str);if(str[0]=='D'){scanf("%d",&x);s.push(x);update(1,1,n,x,0);}else if(str[0]=='Q'){scanf("%d",&x);printf("%d\n",query(1,1,n,x));}else{update(1,1,n,s.top(),1);s.pop();}}}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/chen9510/p/6568385.html

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