POJ Topic 2892 Tunnel Warfare (segment tree single point update query, the maximum continuous interval length of the single point)

Tunnel Warfare

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 7307		Accepted: 2997

Description

During the War of Resistance against Japan, tunnel warfare was carried off extensively in the vast areas of the North China Pl Ain. Generally speaking, villages connected by tunnels lay in a line. Except the ends, every village was directly connected with the 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 is severely isolated, restoration of connection must is done immediately!

Input

The first line of the input contains the positive integers n and m (n, m ≤ 50 , indicating the number of villages and events. Each of the nextm lines describes an event.

There is three different events described in different format shown below:

1. D x: The x-th village was destroyed.
2. Q x: The Army commands requested the number of villages that x-th Village is directly or indirectly con Nected with including itself.
3. R: The village destroyed was rebuilt.

Output

Output 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

Hint

An illustration of the sample input:

      Oooooood 3   ooxooood 6   ooxooxod 5   ooxoxxor     ooxooxor     ooxoooo

Source

POJ monthly--2006.07.30, Updog

AC Code

#include <stdio.h> #include <string.h> #define MAX (A, B) (a>b?a:b) struct s{int rl,ll,ml;} Node[50500<<2];int stack[50500],top;void Build (int l,int r,int tr) {node[tr].ll=node[tr].rl=node[tr].ml=r-l+1; if (l==r) Return;int mid= (l+r) >>1;build (l,mid,tr<<1); build (mid+1,r,tr<<1|1);} void update (int pos,int l,int r,int tr,int val) {if (l==r) {if (val) node[tr].ll=node[tr].rl=node[tr].ml=1;elsenode[tr]. Ll=node[tr].rl=node[tr].ml=0;return;} int mid= (L+R) >>1;if (pos<=mid) {update (pos,l,mid,tr<<1,val);} Elseupdate (Pos,mid+1,r,tr<<1|1,val);node[tr].ll=node[tr<<1].ll;node[tr].rl=node[tr<<1|1].rl; Node[tr].ml=max (NODE[TR&LT;&LT;1].ML,NODE[TR&LT;&LT;1|1].ML); Node[tr].ml=max (node[tr].ml,node[tr<<1].rl+ NODE[TR&LT;&LT;1|1].LL); if (node[tr<<1].ll==mid-l+1) node[tr].ll+=node[tr<<1|1].ll;if (node[tr< <1|1].rl==r-(mid+1) +1) Node[tr].rl+=node[tr<<1].rl;} int query (int pos,int l,int r,int tr) {if (l==r| | node[tr].ml==0| | node[tr].ml==r-l+1) {return node[tr].ml;} int mid= (L+R) >>1;if (pos<=mid) {if (pos>=mid-node[tr<<1].rl+1) return query (pos,l,mid,tr<<1) +query (mid+1,mid+1,r,tr<<1|1); return query (pos,l,mid,tr<<1);} Else{if (POS&LT;=MID+1+NODE[TR&LT;&LT;1|1].LL-1) return query (pos,mid+1,r,tr<<1|1) +query (mid,l,mid,tr< <1); return query (POS,MID+1,R,TR&LT;&LT;1|1);}} int main () {int n,m;while (scanf ("%d%d", &n,&m)!=eof) {build (1,n,1); Top=0;while (m--) {char s[2];int x;scanf ("%s ", s), if (s[0]== ' D ') {scanf ("%d ", &x); Stack[top++]=x;update (x,1,n,1,0);} Else{if (s[0]== ' Q ') {int x;scanf ("%d", &x), int ans=query (x,1,n,1);p rintf ("%d\n", ans);} Else{int x=stack[top-1];top--;update (x,1,n,1,1);}}}}

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POJ Topic 2892 Tunnel Warfare (segment tree single point update query, the maximum continuous interval length of the single point)