Tunnel Warfare 无脑模拟

 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

1. 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

1024

题意： 给你n个桩，m个操作
接下来m行 每行两个数据a b
a=Q时为查询包含b的最大连续子串；
这题是在学长给的线段树练习题中看到的。 看了很久都不会用线段树写，果断无脑暴力模拟一波。

#include <iostream>#include <set>#include <functional>#include<algorithm>#include<cstdio>#include<stack>using namespace std;set<int > S;int main(){    int n,m;    while(cin>>n>>m)    {        int per;        S.clear();        S.insert(0);        S.insert(n+1);        stack<int> ST;        for(int i=0;i<m;i++)        {            int cao;            char dos;            cin>>dos;            if(dos=='D')            {                scanf("%d",&cao);                ST.push(cao);                S.insert(cao);            }            else if(dos=='Q')            {                cin>>cao;                if(S.find(cao)!=S.end())                    puts("0");                else                {                    set<int>::iterator it1=S.upper_bound(cao);                    set<int>::iterator it2=(S.lower_bound(cao));                    it2--;                    //cout<<"it1 : it2 "<<*it1<<" "<<*it2<<endl;                    cout<<(*it1-*it2-1)<<endl;                }            }            else{                S.erase(ST.top());                ST.pop();            }        }    }}