HDU1540 区间合并

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8115    Accepted Submission(s): 3142

Problem 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.

 

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

 

Source

POJ Monthly

 题意：

n个村庄连在一起，最初都完好，D a表示a村庄被破坏，Q a表示询问与a村庄直接或间接相连的有几个，R表示修复了最后破坏的一个村庄。

代码：

//线段树维护pre最大前缀，suf最大后缀，最后求某一点的答案就是其
//前缀加上后缀。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
const int maxn=50004;
int suf[maxn*4],pre[maxn*4];
void pushup(int i,int l,int r)
{
    int m=(l+r)>>1;
    pre[i]=pre[i<<1];
    if(pre[i]==m-l+1) pre[i]+=pre[i<<1|1];
    suf[i]=suf[i<<1|1];
    if(suf[i]==r-m) suf[i]+=suf[i<<1];
}
void build(int i,int l,int r)
{
    if(l==r){
        suf[i]=pre[i]=1;
        return ;
    }
    int m=(l+r)>>1;
    build(i<<1,l,m);
    build(i<<1|1,m+1,r);
    pushup(i,l,r);
}
void update(int id,int c,int i,int l,int r)
{
    if(l==r){
        suf[i]=pre[i]=c;
        return ;
    }
    int m=(l+r)>>1;
    if(id<=m) update(id,c,i<<1,l,m);
    else update(id,c,i<<1|1,m+1,r);
    pushup(i,l,r);
}
int query_pre(int ql,int qr,int i,int l,int r)
{
    if(ql<=l&&qr>=r) return pre[i];
    int m=(l+r)>>1,res;
    if(qr<=m) return query_pre(ql,qr,i<<1,l,m);
    if(ql>m) return query_pre(ql,qr,i<<1|1,m+1,r);
    res=query_pre(ql,qr,i<<1,l,m);
    if(res==m-max(l,ql)+1) res+=query_pre(ql,qr,i<<1|1,m+1,r);
    //注意记住写上max函数
    return res;
}
int query_suf(int ql,int qr,int i,int l,int r)
{
    if(ql<=l&&qr>=r) return suf[i];
    int m=(l+r)>>1,res;
    if(qr<=m) return query_suf(ql,qr,i<<1,l,m);
    if(ql>m) return query_suf(ql,qr,i<<1|1,m+1,r);
    res=query_suf(ql,qr,i<<1|1,m+1,r);
    if(res==min(r,qr)-m) res+=query_suf(ql,qr,i<<1,l,m);
    return res;
}
int main()
{
    int n,m,x;
    char ch[3];
    while(scanf("%d%d",&n,&m)==2){
        build(1,1,n);
        stack<int>s;
        while(m--){
            scanf("%s",ch);
            if(ch[0]=='D'){
                scanf("%d",&x);
                update(x,0,1,1,n);
                s.push(x);
            }
            else if(ch[0]=='R'){
                if(!s.empty()){
                    int x=s.top();s.pop();
                    update(x,1,1,1,n);
                }
            }
            else if(ch[0]=='Q'){
                scanf("%d",&x);
                int ans=query_suf(1,x,1,1,n);
                if(ans==0) printf("%d
",ans);
                else{
                    ans+=query_pre(x,n,1,1,n)-1;
                    printf("%d
",ans);
                }
            }
        }
    }
    return 0;
}