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poj 2892---Tunnel Warfare（线段树单点更新、…

2018-06-17 23:04:39来源：未知阅读 ()

题目链接

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

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

题意：有n个村庄编号为1，2，3...n 它们按照序号一一相连，现在有m次操作，有以下几种操作：

1、D x 表示将x号村庄摧毁。

2、Q x 表示查询x村庄能到达的村庄数（包括x村庄）。

3、R 表示修复最近一个被摧毁的村庄。

每次查询输出一个值。

思路：线段树单点更新、区间合并，用栈存储被摧毁的村庄号。

代码如下：

#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;
}