Tunnel Warfare
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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
D 3
D 6
D 5
Q 4
Q 5
R
Q 4
R
Q 4
Sample Output
1
0
2
4
0
2
4
Source
题解：区间合并基础题
//
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
#include<stack>
///#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define memfy(a) memset(a,-1,sizeof(a))
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b) scanf("%d%d",&a,&b)
#define mod 530600414
#define eps 0.0000000001
#define inf 1000000000.0
inline ll read()
{
ll x=,f=;
char ch=getchar();
while(ch<''||ch>'')
{
if(ch=='-')f=-;
ch=getchar();
}
while(ch>=''&&ch<='')
{
x=x*+ch-'';
ch=getchar();
}
return x*f;
}
//**************************************** #define maxn 500000+5
struct ss
{
int l,r,L,R,v;
}tr[maxn];
int n,m,lflag,rflag;
void pushup(int k)
{
if(tr[k<<].L+tr[k<<|].R==(tr[k].r-tr[k].l+))
{
tr[k].L=tr[k].R=(tr[k].r-tr[k].l+);
}
else {
if(tr[k<<].L>=(tr[k<<].r-tr[k<<].l+))
tr[k].L=tr[k<<].L+tr[k<<|].L;
else tr[k].L=tr[k<<].L;
if(tr[k<<|].R>=tr[k<<|].r-tr[k<<|].l+)
tr[k].R=tr[k<<|].R+tr[k<<].R;
else tr[k].R=tr[k<<|].R;
}
}
void build(int k,int s,int t)
{
tr[k].l=s;
tr[k].r=t;
tr[k].L=tr[k].R=t-s+;
if(s==t)
{
return ;
}
int mid=(s+t)>>;
build(k<<,s,mid);
build(k<<|,mid+,t);
}
void update(int k,int x,int c)
{
if(tr[k].l==x&&tr[k].r==x)
{
tr[k].L=tr[k].R=c;return ;
}
int mid=(tr[k].l+tr[k].r)>>;
if(x<=mid)update(k<<,x,c);
else update(k<<|,x,c);
pushup(k);
}
int ask(int k,int x)
{
if(tr[k].l==tr[k].r&&tr[k].l==x)
{
if(tr[k].L)rflag=lflag=;
return tr[k].L;
}
int mid=(tr[k].l+tr[k].r)>>;
int A=,B=;
if(x<=mid)
{
A=ask(k<<,x);
}
else {
B=ask(k<<|,x);
}
int ans=A+B;
if(A && rflag)
{
ans+=tr[k<<|].L;
if(tr[k<<|].L<tr[k<<|].r-tr[k<<|].l+)
rflag=;
}
if(B && lflag)
{
ans+=tr[k<<].R;
if(tr[k<<].R<tr[k<<].r-tr[k<<].l+)
lflag=;
}
return ans;
}
int main()
{
char ch[];
int x; while(scanf("%d%d",&n,&m)!=EOF)
{
stack<int >q;
int last=;
build(,,n);
FOR(i,,m)
{
scanf("%s",ch);
if(ch[]=='D')
{cin>>x;
q.push(x);
update(,x,);
}
else if(ch[]=='Q')
{
cin>>x;
rflag=lflag=;
cout<<ask(,x)<<endl;
}
else {
last=q.top();
q.pop();
update(,last,);
}
}
}
return ;
}
代码
补个 treap写法
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
struct data{
int l,r,v,size,rnd,w;
}tr[];
int n,size,root,ans;
int last[];
void update(int k)//更新结点信息
{
tr[k].size=tr[tr[k].l].size+tr[tr[k].r].size+tr[k].w;
}
void rturn(int &k)
{
int t=tr[k].l;tr[k].l=tr[t].r;tr[t].r=k;
tr[t].size=tr[k].size;update(k);k=t;
}
void lturn(int &k)
{
int t=tr[k].r;tr[k].r=tr[t].l;tr[t].l=k;
tr[t].size=tr[k].size;update(k);k=t;
}
void insert(int &k,int x)
{
if(k==)
{
size++;k=size;
tr[k].size=tr[k].w=;tr[k].v=x;tr[k].rnd=rand();
return;
}
tr[k].size++;
if(tr[k].v==x)tr[k].w++;
else if(x>tr[k].v)
{
insert(tr[k].r,x);
if(tr[tr[k].r].rnd<tr[k].rnd)lturn(k);
}
else
{
insert(tr[k].l,x);
if(tr[tr[k].l].rnd<tr[k].rnd)rturn(k);
}
}
void del(int &k,int x)
{
if(k==)return;
if(tr[k].v==x)
{
if(tr[k].w>)
{
tr[k].w--;tr[k].size--;return;
}
if(tr[k].l*tr[k].r==)k=tr[k].l+tr[k].r;
else if(tr[tr[k].l].rnd<tr[tr[k].r].rnd)
rturn(k),del(k,x);
else lturn(k),del(k,x);
}
else if(x>tr[k].v)
tr[k].size--,del(tr[k].r,x);
else tr[k].size--,del(tr[k].l,x);
} int query_rank(int k,int x)
{
if(k==)return ;
if(tr[k].v==x)return tr[tr[k].l].size+;
else if(x>tr[k].v)
return tr[tr[k].l].size+tr[k].w+query_rank(tr[k].r,x);
else return query_rank(tr[k].l,x);
} int query_num(int k,int x)
{
if(k==)return ;
if(x<=tr[tr[k].l].size)
return query_num(tr[k].l,x);
else if(x>tr[tr[k].l].size+tr[k].w)
return query_num(tr[k].r,x-tr[tr[k].l].size-tr[k].w);
else return tr[k].v;
} void query_pre(int k,int x)
{
if(k==)return;
if(tr[k].v == x) {
ans = k;
return ;
}
if(tr[k].v<x)
{
ans=k;query_pre(tr[k].r,x);
}
else query_pre(tr[k].l,x);
}
void query_sub(int k,int x)
{
if(k==)return;
if(tr[k].v == x) {
ans = k;
return ;
}
if(tr[k].v>x)
{
ans=k;query_sub(tr[k].l,x);
}
else query_sub(tr[k].r,x);
}
int m;
map<int,int > mp;
int main()
{
while(scanf("%d%d",&n,&m)!=EOF) {
int cnt = ,t1, t2 , x;
root = ;
mp.clear();
tr[root].size = ;
char ch[];
insert(root,);
insert(root,n+);
for(int i=;i<=m;i++) {
scanf("%s",ch);
if(ch[]=='D') {
scanf("%d",&x) ;
if(mp[x]==) {
insert(root,x);
mp[x] = ;
}last[++cnt] = x; }
if(ch[]=='R') {
if(!cnt) continue;
if(mp[last[cnt]] == ) {
del(root,last[cnt]);
mp[last[cnt]] = ;
}
cnt--;
}
if(ch[]=='Q') {
scanf("%d",&x);
ans = ;
query_sub(root,x);
t1= tr[ans].v ;
ans = ;
query_pre(root,x) ;
t2 = tr[ans].v;
printf("%d\n", max(t1 - t2 - ,));
}
}
}
}