Tunnel Warfare
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=1540
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
题意
给你一个区间，有三个操作，使的一个村庄毁灭，使的上一个毁灭的村庄复活，查询这个村庄所在最长区间
题解：
啊，就是一个区间合并的简单线段树，单点修改，记录从左边的最大值，右边的最大值，区间的最大值
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 150001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff; //无限大
const int inf=0x3f3f3f3f;
/*
int buf[10];
inline void write(int i) {
int p = 0;if(i == 0) p++;
else while(i) {buf[p++] = i % 10;i /= 10;}
for(int j = p-1; j >=0; j--) putchar(‘0‘ + buf[j]);
printf("\n");
}
*/
//**************************************************************************************
inline ll read(){
int x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
struct node{
int l,r;
int len;
int lm,rm,mm;
};
node a[maxn*4];
void build(int l,int r,int x){
a[x].l=l,a[x].r=r;
a[x].lm=a[x].rm=a[x].len=a[x].mm=r-l+1;
if(l!=r){
int mid=(l+r)>>1;
build(l,mid,x<<1);
build(mid+1,r,x<<1|1);
}
}
void updata(int i,int t,int x)
{
if(a[i].l == a[i].r){a[i].lm = a[i].rm = a[i].mm = x;return;}
int mid = (a[i].l+a[i].r)>>1;
if(t<=mid)updata(2*i,t,x);
else updata(2*i+1,t,x);
a[i].lm = a[2*i].lm;
a[i].rm = a[2*i+1].rm;
a[i].mm = max(max(a[2*i].mm,a[2*i+1].mm),a[2*i].rm+a[2*i+1].lm);
if(a[2*i].lm == a[2*i].r-a[2*i].l+1)a[i].lm += a[2*i+1].lm;
if(a[2*i+1].rm == a[2*i+1].r-a[2*i+1].l+1)a[i].rm += a[2*i].rm;
}
int query(int x,int p)
{
if(a[x].l==a[x].r||a[x].mm==0||a[x].mm==a[x].len)
return a[x].mm;
int mid=(a[x].l+a[x].r)>>1;
if(p<=mid){
if(p>=a[x<<1].r-a[x<<1].rm+1)
return query(x<<1,p)+query(x<<1|1,mid+1);
else
return query(x<<1,p);
}
else{
if(p<=a[x<<1|1].l+a[x<<1|1].lm-1)
return query(x<<1|1,p)+query(x<<1,mid);
else
return query(x<<1|1,p);
}
}
int main()
{
int n,m,y;
char x[3];
while(scanf("%d%d",&n,&m)!=EOF)
{
stack<int> k;
memset(a,0,sizeof(a));
build(1,n,1);
for(int i=0;i<m;i++)
{
scanf("%s",x);
if(x[0]==‘D‘)
{
y=read();
updata(1,y,0);
k.push(y);
}
else if(x[0]==‘Q‘)
{
y=read();
printf("%d\n",query(1,y));
}
else
{
updata(1,k.top(),1);
k.pop();
}
}
}
}
