Tunnel Warfare

http://acm.hdu.edu.cn/showproblem.php?pid=1540

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13440 Accepted Submission(s): 5333

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

Q 4

Sample Output

Source

POJ Monthly

注意是多組數據。

這是用最大值最小值單點更新的方法，參考大佬的博客 https://blog.csdn.net/chudongfang2015/article/details/52133243

一開始設所有村庄最大值為0，最小值為n+1.

當村庄被摧毀時，它的最大值和最小值設為改村庄的編號，這樣我們從左邊區間查詢最大值max，右邊區間查詢最小值min

當min==max時，就是該點被摧毀了，否則區間長度就是min-max-1

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<algorithm>
  5 #include<queue>
  6 #include<vector>
  7 #include<string>
  8 #include<stack>
  9 #define maxn 50005
 10 #define lson l,mid,rt<<1
 11 #define rson mid+1,r,rt<<1|1
 12 using namespace std;
 13 
 14 int n,m;
 15 struct sair{
 16     int Max,Min;
 17 }tree[maxn<<3];
 18 
 19 void build(int l,int r,int rt){
 20     if(l==r){
 21         tree[rt].Max=0;
 22         tree[rt].Min=n+1;
 23         return;
 24     }
 25     int mid=(l+r)/2;
 26     build(lson);
 27     build(rson);
 28     tree[rt].Max=max(tree[rt<<1].Max,tree[rt<<1|1].Max);
 29     tree[rt].Min=min(tree[rt<<1].Min,tree[rt<<1|1].Min);
 30 
 31 }
 32 
 33 void update_max(int L,int k,int l,int r,int rt){
 34     if(l==r){
 35         tree[rt].Max=k;
 36         return;
 37     }
 38     int mid=(l+r)/2;
 39     if(L<=mid) update_max(L,k,lson);
 40     else update_max(L,k,rson);
 41     tree[rt].Max=max(tree[rt<<1].Max,tree[rt<<1|1].Max);
 42 }
 43 
 44 void update_min(int L,int k,int l,int r,int rt){
 45     if(l==r){
 46         tree[rt].Min=k;
 47         return;
 48     }
 49     int mid=(l+r)/2;
 50     if(L<=mid) update_min(L,k,lson);
 51     else update_min(L,k,rson);
 52     tree[rt].Min=min(tree[rt<<1].Min,tree[rt<<1|1].Min);
 53 }
 54 
 55 int query_max(int L,int R,int l,int r,int rt){
 56     if(L<=l&&R>=r){
 57         return tree[rt].Max;
 58 
 59     }
 60     int mid=(l+r)/2;
 61     int ans=0;
 62     if(L<=mid) ans=max(ans,query_max(L,R,lson));
 63     if(R>mid) ans=max(ans,query_max(L,R,rson));
 64     return ans;
 65 }
 66 
 67 int query_min(int L,int R,int l,int r,int rt){
 68     if(L<=l&&R>=r){
 69         return tree[rt].Min;
 70     }
 71     int mid=(l+r)/2;
 72     int ans=0x3f3f3f3f;
 73     if(L<=mid) ans=min(ans,query_min(L,R,lson));
 74     if(R>mid) ans=min(ans,query_min(L,R,rson));
 75     return ans;
 76 }
 77 
 78 int main(){
 79 
 80     std::ios::sync_with_stdio(false);
 81     while(cin>>n>>m){
 82         char pos;
 83         int x;
 84         stack<int>st;
 85         build(1,n,1);
 86         for(int i=1;i<=m;i++){
 87             cin>>pos;
 88             if(pos=='D'){
 89                 cin>>x;
 90                 st.push(x);
 91                 update_max(x,x,1,n,1);
 92                 update_min(x,x,1,n,1);
 93             }
 94             else if(pos=='Q'){
 95                 cin>>x;
 96                 int L=query_min(x,n,1,n,1);
 97                 int R=query_max(1,x,1,n,1);
 98                 if(R==L) cout<<0<<endl;
 99                 else cout<<L-R-1<<endl;
100             }
101             else if(pos=='R'){
102                 x=st.top();
103                 st.pop();
104                 update_max(x,0,1,n,1);
105                 update_min(x,n+1,1,n,1);
106             }
107         }
108     }
109 }

View Code

區間合並

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <string>
 4 #include <cmath>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <queue>
 8 #include <stack>
 9 #include <vector>
10 #include <set>
11 #include <map>
12 #define maxn 50010
13 #define lson l,mid,rt<<1
14 #define rson mid+1,r,rt<<1|1
15 using namespace std;
16 
17 int tree[maxn<<3],lsum[maxn<<3],rsum[maxn<<3];
18 //總數，左節點向右的連續個數，右節點向左的連續個數
19 
20 void pushup(int len,int rt){
21     lsum[rt]=lsum[rt<<1];
22     if(lsum[rt]==(len-len/2)) lsum[rt]+=lsum[rt<<1|1];
23     rsum[rt]=rsum[rt<<1|1];
24     if(rsum[rt]==len/2) rsum[rt]+=rsum[rt<<1];
25     tree[rt]=max(lsum[rt<<1|1]+rsum[rt<<1],max(tree[rt<<1],tree[rt<<1|1]));
26 }
27 
28 void build(int l,int r,int rt){
29     if(l==r){
30         tree[rt]=lsum[rt]=rsum[rt]=1;
31         return;
32     }
33     int mid=(l+r)/2;
34     build(lson);
35     build(rson);
36     pushup(r-l+1,rt);
37 }
38 
39 void add(int L,int k,int l,int r,int rt){
40     if(l==r){
41         tree[rt]=lsum[rt]=rsum[rt]=k;
42         return;
43     }
44     int mid=(l+r)/2;
45     if(L<=mid) add(L,k,lson);
46     else add(L,k,rson);
47     pushup(r-l+1,rt);
48 }
49 
50 int query(int L,int l,int r,int rt){
51     if(l==r) return tree[rt];
52     int mid=(l+r)/2;
53     if(L<=mid){
54         if(L+rsum[rt<<1]>mid) return rsum[rt<<1]+lsum[rt<<1|1];
55         //查詢該點是否在范圍內
56         return query(L,lson);
57     }
58     else{
59         if(mid+lsum[rt<<1|1]>=L) return rsum[rt<<1]+lsum[rt<<1|1];
60         return query(L,rson);
61     }
62 }
63 
64 int main(){
65     std::ios::sync_with_stdio(false);
66     int n,m,x;
67     string pos;
68     while(cin>>n>>m){
69         stack<int>st;
70         build(1,n,1);
71 
72         for(int i=1;i<=m;i++){
73             cin>>pos;
74             if(pos[0]=='Q'){
75                 cin>>x;
76                 cout<<query(x,1,n,1)<<endl;
77             }
78             else if(pos[0]=='D'){
79                 cin>>x;
80                 st.push(x);
81                 add(x,0,1,n,1);
82             }
83             else{
84                 x=st.top();
85                 st.pop();
86                 add(x,1,1,n,1);
87             }
88         }
89     }
90 
91 }
92

View Code

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