HDU

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
 
题意：
操作1：从x位置往后插y枝花，前提是只能给空花瓶插，并求出最左边和最右边插花的位置。
操作2：求[x, y]内有多少只花，并清空区间内的花瓶。
题解：
因为每个花瓶只能插一枝花，所以用区间和可以判断能插几只花。
对于操作1，我们可以判断[x, n]的空花瓶数cnt，如果为0，直接GG；否则看它能插最多y枝花还是cnt枝花。
通过二分可以找到这个l, r的位置；
对于操作2，就是区间求和，区间更新，裸的。
代码：
  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 }
View Code
 
Author
【推广】 免费学中医，健康全家人
原文地址：https://www.cnblogs.com/Jstyle-continue/p/6396816.html