Tunnel Warfare POJ - 2892（SET）

最新推荐文章于 2021-01-23 23:18:45 发布

原创最新推荐文章于 2021-01-23 23:18:45 发布 · 380 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#poj #set

acm 同时被 2 个专栏收录

24 篇文章

订阅专栏

算法学习

18 篇文章

订阅专栏

本文介绍了一个模拟抗日战争期间地道战的算法模型。该模型通过一系列输入事件（如村庄被破坏、重建或查询连接状态），利用数据结构高效地计算并输出各村庄间的连接状态。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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
Hint
An illustration of the sample input:

  OOOOOOO

D 3 OOXOOOO

D 6 OOXOOXO

D 5 OOXOXXO

R OOXOOXO

R OOXOOOO

查询每个点的连通数即与他相连的O的数量+1，只需要找到其往左第一个X和往右第一个X，然后求差值-1，注意边界。用set处理很方便。

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<string>
#include<map>
#include<set>
#include<vector>
using namespace std;
bool p[50005];
int d[50005];
int n,m;
int dnum;
set<int> s;
int main(){
    scanf("%d%d",&n,&m);
    dnum=0;
    string str;
    int num;
    for(int i=1;i<=n;i++){p[i]=1;}
    for(int i=0;i<m;i++){
        cin>>str;
        if(str=="D"){scanf("%d",&num);
            d[dnum++]=num;
            s.insert(num);
            p[num]=0;
        }
        else if(str=="Q"){
            scanf("%d",&num);
            if(p[num]==0){printf("0\n");}
            else{
                set<int>::iterator it=s.lower_bound(num);
                int ans=n+1;
                if(it!=s.end()){ans=(*it);}
                if(it!=s.begin()){
                    it--;
                    ans-=(*it);
                }
                ans--;
                printf("%d\n",ans);
            }
        }
        else{
            dnum--;
            p[d[dnum]]=1;
            s.erase(s.find(d[dnum]));
        }
    }
return 0;
}