Tunnel WarfareTime Limit:1000MSMemory Limit:131072KTotal Submissions:7434Accepted:3070

    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 integersnandm(n,m≤ 50,000) indicating the number of villages and events. Each of the nextmlines describes an event.

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

D x: Thex-th village was destroyed.Q x: The Army commands requested the number of villages thatx-th village was directly or indirectly connected with including itself.R: The village destroyed last 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

    平衡樹的應(yīng)用

    平衡樹中保存所有被炸毀的節(jié)點(diǎn)，

poj2892 Tunnel Warface

，

電腦資料

《poj2892 Tunnel Warface》(http://www.oriental01.com)。

    對(duì)于炸毀和修復(fù)操作，直接在平衡樹中插入或刪除。

    對(duì)于查詢操作，先判斷該節(jié)點(diǎn)是否被炸毀，如果被炸毀答案為0，否則在平衡樹中求出前驅(qū)x和后繼y，答案為y-x-1。

    還有一種方法是二分+樹狀數(shù)組，感覺速度比這個(gè)慢就沒有寫…

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>#include#include<stack>#define F(i,j,n) for(int i=j;i<=n;i++)#define D(i,j,n) for(int i=j;i>=n;i--)#define LL long long#define MAXN 50005#define pa pair<int,int>#define INF 1000000000using namespace std;int n,m,x,tot=0,rt=0,l[MAXN],r[MAXN],rnd[MAXN],v[MAXN];char op;bool f[MAXN];stack<int>st;inline int read(){	int ret=0,flag=1;char ch=getchar();	while (ch<'0'||ch>'9'){if (ch=='-') flag=-1;ch=getchar();}	while (ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}	return ret*flag;}inline void rturn(int &k){	int tmp=l[k];	l[k]=r[tmp];r[tmp]=k;	k=tmp;}inline void lturn(int &k){	int tmp=r[k];	r[k]=l[tmp];l[tmp]=k;	k=tmp;}inline void ins(int &k,int x){	if (!k){k=++tot;v[k]=x;l[k]=r[k]=0;rnd[k]=rand();return;}	if (x<v[k]) else="" if="" inline="" int="" k="l[k]+r[k];" return="" tmp="pre(r[k],x);return" void="" x="">=v[k]) return suc(r[k],x);	else {int tmp=suc(l[k],x);return tmp==n+1?v[k]:tmp;}}inline int getans(int x){	if (f[x]) return 0;	return suc(rt,x)-pre(rt,x)-1;}int main(){	memset(f,false,sizeof(f));	n=read();m=read();	while (m--)	{		p=getchar();while (op<'A'||op>'Z') p=getchar();		if (op=='D') {x=read();f[x]=true;st.push(x);ins(rt,x);}		else if (op=='Q') {x=read();printf("%d\n",getans(x));}		else {del(rt,st.top());f[st.top()]=false;st.pop();}	}}</v[k])></int></int,int></stack></cstdlib></cstring></cmath></cstdio></iostream>