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平衡树的应用
平衡树中保存所有被炸毁的节点，
poj2892 Tunnel Warface
，电脑资料
《poj2892 Tunnel Warface》(https://www.unjs.com)。对于炸毁和修复操作，直接在平衡树中插入或删除。
对于查询操作，先判断该节点是否被炸毁，如果被炸毁答案为0，否则在平衡树中求出前驱x和后继y，答案为y-x-1。
还有一种方法是二分+树状数组，感觉速度比这个慢就没有写…
#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>