HDU - 4614 【二分+線段樹維護】

Vases and Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3263    Accepted Submission(s): 1299

Problem Description

　　Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them in the vases, one flower in one vase. She randomly choose the vase A and try to put a flower in the vase. If the there is no flower in the vase, she will put a flower in it, otherwise she skip this vase. And then she will try put in the vase A+1, A+2, ..., N-1, until there is no flower left or she has tried the vase N-1. The left flowers will be discarded. Of course, sometimes she will clean the vases. Because there are too many vases, she randomly choose to clean the vases numbered from A to B(A <= B). The flowers in the cleaned vases will be discarded.

 

 

Input

　　The first line contains an integer T, indicating the number of test cases.
　　For each test case, the first line contains two integers N(1 < N < 50001) and M(1 < M < 50001). N is the number of vases, and M is the operations of Alice. Each of the next M lines contains three integers. The first integer of one line is K(1 or 2). If K is 1, then two integers A and F follow. It means Alice receive F flowers and try to put a flower in the vase A first. If K is 2, then two integers A and B follow. It means the owner would like to clean the vases numbered from A to B(A <= B).

 

 

Output

　　For each operation of which K is 1, output the position of the vase in which Alice put the first flower and last one, separated by a blank. If she can not put any one, then output 'Can not put any one.'. For each operation of which K is 2, output the number of discarded flowers.
　　 Output one blank line after each test case.

 

 

Sample Input

2 10 5 1 3 5 2 4 5 1 1 8 2 3 6 1 8 8 10 6 1 2 5 2 3 4 1 0 8 2 2 5 1 4 4 1 2 3

 

 

Sample Output

3 7

2

1 9

4

Can not put any one.

 

2 6

2

0 9

4

4 5

2 3

 

題意：

操做1：從x位置日後插y枝花，前提是隻能給空花瓶插，並求出最左邊和最右邊插花的位置。

操做2：求[x, y]內有多少隻花，並清空區間內的花瓶。

題解：

由於每一個花瓶只能插一枝花，因此用區間和能夠判斷能插幾隻花。

對於操做1，咱們能夠判斷[x, n]的空花瓶數cnt，若是爲0，直接GG；不然看它能插最多y枝花仍是cnt枝花。

經過二分能夠找到這個l, r的位置；

對於操做2，就是區間求和，區間更新，裸的。

代碼：

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <cstring>
  4 #include <cstdio>
  5 #include <bitset>
  6 #include <vector>
  7 #include <queue>
  8 #include <stack>
  9 #include <cmath>
 10 #include <list>
 11 #include <set>
 12 #include <map>
 13 #define rep(i,a,b) for(int i = a;i <= b;++ i)
 14 #define per(i,a,b) for(int i = a;i >= b;-- i)
 15 #define mem(a,b) memset((a),(b),sizeof((a)))
 16 #define FIN freopen("in.txt","r",stdin)
 17 #define FOUT freopen("out.txt","w",stdout)
 18 #define IO ios_base::sync_with_stdio(0),cin.tie(0)
 19 #define mid ((l+r)>>1)
 20 #define ls (id<<1)
 21 #define rs ((id<<1)|1)
 22 using namespace std;
 23 typedef long long LL;
 24 typedef pair<int, int> PIR;
 25 const int N = 50005;
 26 struct Node{
 27     int sum, lazy;
 28 }node[N*4];
 29 int T, n, m, op, x, y;
 30 
 31 void pushUp(int id, int l, int r){
 32     node[id].sum = node[ls].sum+node[rs].sum;
 33 }
 34 void pushDown(int id, int l, int r){
 35     node[ls].sum = node[id].lazy*(mid-l+1);
 36     node[rs].sum = node[id].lazy*(r-mid);
 37     node[ls].lazy = node[id].lazy;
 38     node[rs].lazy = node[id].lazy;
 39     node[id].lazy = -1;
 40 }
 41 void build(int id, int l, int r){
 42     node[id].lazy = -1;
 43     if(l == r){
 44         node[id].sum = 0;
 45         return ;
 46     }
 47     build(ls, l, mid);
 48     build(rs, mid+1, r);
 49     pushUp(id, l, r);
 50 }
 51 void update(int id, int l, int r, int ql, int qr, int p){
 52     if(l == ql && r == qr){
 53         node[id].lazy = p;
 54         node[id].sum = (r-l+1)*p;
 55         return ;
 56     }
 57     if(node[id].lazy != -1)
 58         pushDown(id, l, r);
 59     if(qr <= mid)   update(ls, l, mid, ql, qr, p);
 60     else if(ql > mid)
 61         update(rs, mid+1, r, ql, qr, p);
 62     else{
 63         update(ls, l, mid, ql, mid, p);
 64         update(rs, mid+1, r, mid+1, qr, p);
 65     }
 66     pushUp(id, l, r);
 67 }
 68 int query(int id, int l, int r, int ql, int qr){
 69     if(l == ql && r == qr){
 70         return node[id].sum;
 71     }
 72     if(node[id].lazy != -1)   pushDown(id, l, r);
 73     if(qr <= mid)   return query(ls, l, mid, ql, qr);
 74     else if(ql > mid)
 75         return query(rs, mid+1, r, ql, qr);
 76     else{
 77         return query(ls, l, mid, ql, mid)+query(rs, mid+1, r, mid+1, qr);
 78     }
 79 }
 80 int main()
 81 {IO;
 82     //FIN;
 83     cin >> T;
 84     while(T--){
 85         cin >> n >> m;
 86         build(1, 1, n);
 87         rep(i, 1, m){
 88             cin >> op >> x >> y;
 89             if(op == 1){
 90                 x++;
 91                 int cnt = n-x+1-query(1, 1, n, x, n);
 92                 if(cnt == 0){
 93                     cout << "Can not put any one." << endl;
 94                     continue;
 95                 }
 96                 cnt = min(cnt, y);
 97                 int low = x, high = n, ansl, ansr;
 98                 while(low <= high){
 99                     int md = (low+high)>>1;
100                     if(query(1, 1, n, x, md) < md-x+1){
101                         ansl = md;
102                         high = md-1;
103                     }
104                     else{
105                         low = md+1;
106                     }
107                 }
108                 low = x, high = n;
109                 while(low <= high){
110                     int md = (low+high)>>1;
111                     if(md-x+1-query(1, 1, n, x, md) >= cnt){
112                         ansr = md;
113                         high = md-1;
114                     }
115                     else{
116                         low = md+1;
117                     }
118                 }
119                 update(1, 1, n, ansl, ansr, 1);
120                 cout << ansl-1 << " " << ansr-1 << endl;
121             }
122             else{
123                 x++;
124                 y++;
125                 int ans = query(1, 1, n, x, y);
126                 update(1, 1, n, x, y, 0);
127                 cout << ans << endl;
128             }
129         }
130         cout << endl;
131     }
132     return 0;
133 }

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