topic:
Winter came, and went to Xiao Ming and goddesses dating season.
Xiao Ming Although the grass root level code farmers, but very active, enthusiastic goddess who often reply "Oh," following a statement by a large segment of the Internet Xiaoming, so, Xiao Ming is a favorite and goddesses dating. At the same time, there are many gay friend asked him to open the black, because the number is too large, how to arrange a time Xiao Ming has become a major mind.
There will be many or goddess-based Friends come to us known Xiao Ming Xiao Ming, a total of T idle time period.
As an operating system once anger exam 71 points Great God, Xiao Ming think of an algorithm, that is, "first fit algorithm", according to the description of the operating system textbooks, it is to find contiguous space for some of the most forward of the meet the requirements assigned to each request, thus Xiao Ming made a decision:
when a gay friend came Xiaoming, according to Xiao Ming on "first-fit algorithm" to find some free time and based Friends of the appointment, if found, say "X, let's fly" (here, X is the start time), otherwise say "fly with yourself";
when the goddess came to Bob, first use a "first fit algorithm", if not found, Xiao Ming would run the risk of ignoring all wood Jiji Cock silk-based Friends of the convention, again using the "ignore-based Friends of the first-fit algorithm", twice as long as one is found, say "X, do not put my gezi " ( here, X is the start time), otherwise say " wait for me "
of course, we know who Bob is not a festival parade negative infinity if and goddesses dating finished, time remaining, and he will be based Friends of the original appointment to dota. (For example: Konishi (Cock wire) Xiaoming appointment play dota within 1 to 5 the unit of time period, at this time, the goddess to Xiaoming reservation length of period 3, the final is 1 to 3, Xiaoming to goddess dating, playing dota at 4-5 and get back Konishi)
Xiao Ming occasionally want to learn new knowledge, at this time Xiao Ming will all have a certain predetermined time interval of time to learn all the empty and Shout "I am the hope of chinese chengxuyuan !!", but are generally Xiaoming heat for three minutes, then it was to reserve it, Xiao Ming will be unable to bear alone the time to learn new knowledge dispensed.
Input
CASE input of the first behavior, expressed CASE set of test data;
each set of data in two integers T, begins N T represents the total time, N is the number of reservation request;
next N rows, each row represents one or goddess based Friends of the appointment, "NS QT" on behalf of a goddess came Xiaoming about for some length of time the QT, "DS QT" represents a Cock wire length request QT, of course, also possible that Bob wanted to learn the knowledge of, "sTUDY !! LR "all requests within the interval representative of emptying L ~ R.
[Technical Specification]
1.. 1 <= the CASE <= 30
2.. 1 <= T, N <= 100000
3.. 1 <= QT <= 110000
4.. 1 <= L <= R & lt <= T
Output
For each Case, the first line of the first output "Case C:" is the first of several representatives Case, then N rows, each row corresponding to a result of the request (see below).
Output samples (can be copied here):
"X-, the let apos Fly", "Fly with Yourself", "X-, do Not My Gezi PUT", "the wait for Me", "The Hope of the I AM chinese chengxuyuan !!"
Sample Input
1 5 6 DS 3 NS 2 NS 4 STUDY!! 1 5 DS 4 NS 2
Sample Output
Case 1: 1,let's fly 4,don't put my gezi wait for me I am the hope of chinese chengxuyuan!! 1,let's fly 1,don't put my gezi
:( code analysis longest since freshman ever seen, are 200+ online, he just took me down to the collapse of the 150+, too much trouble ...) this troublesome question is how to find the range of queries from the left the first free zone. See the specific implementation of code >>
Code:
1 #include <bits / STDC ++. H> ///////////////////// longest ever written code of topics, mygod! 2 the using namespace STD; . 3 const int MAXN 1E5 + = 10 ; . 4 struct Node . 5 { . 6 int L, R & lt; . 7 int RM, mm, LM; //// Cock wire . 8 int RM1, MM1, LM1; //// / goddess . 9 void INI () = mm {RM = LM = 0 ;} 10 void Full () = {RM = R & lt mm LM = - L + . 1 ;} . 11 void INI1 () = {RM1 MM1 = LM1 = 0; } 12 void full1() { rm1 = mm1 = lm1 = r - l + 1; } 13 }t[maxn << 2]; 14 int n, m; 15 16 void pushdown(int tar) 17 { 18 if (t[tar].mm == 0) 19 t[tar << 1].ini(), t[tar << 1 | 1].ini(); 20 if (t[tar].mm == t[tar].r - t[tar].l + 1) 21 t[tar << 1].full(), t[tar << 1 | 1].full(); 22 if(t[tar].mm1 == 0) 23 t[tar << 1].ini1(), t[tar << 1 | 1].ini1(); 24 if (t[tar].mm1 == t[tar].r - t[tar].l + 1) 25 t[tar << 1].full1(), t[tar << 1 | 1].full1(); 26 } 27 28 void pushup(int tar) 29 { 30 t[tar].lm = t[tar << 1].lm; 31 t[tar].rm = t[tar << 1 | 1].rm; 32 t[tar].mm = max(t[tar << 1].mm, t[tar << 1 | 1].mm); 33 t[tar].mm = max(t[tar].mm, t[tar << 1].rm + t[tar << 1 | 1].lm); 34 if (t[tar << 1].lm == t[tar << 1].r - t[tar << 1].l + 1) 35 t[tar].lm += t[tar << 1 | 1].lm; 36 if (t[tar << 1 | 1].rm == t[tar << 1 | 1].r - t[tar << 1 | 1].l + 1) 37 t[tar].rm += t[tar << 1].rm; 38 39 t[tar].lm1 = t[tar << 1].lm1; 40 t[tar].rm1 = t[tar << 1 | 1].rm1; 41 t[tar].mm1 = max(t[tar << 1].mm1, t[tar << 1 | 1].mm1); 42 t[tar].mm1 = max(t[tar].mm1, t[tar << 1].rm1 + t[tar << 1 | 1].lm1); 43 if (t[tar << 1].lm1 == t[tar << 1].r - t[tar << 1].l + 1) 44 t[tar].lm1 += t[tar << 1 | 1].lm1; 45 if (t[tar << 1 | 1].rm1 == t[tar << 1 | 1].r - t[tar << 1 | 1].l + 1) 46 t[tar].rm1 += t[tar << 1].rm1; 47 } 48 49 void build(int l, int r, int tar) 50 { 51 t[tar].l = l, t[tar].r = r; 52 t[tar].full(), t[tar].full1(); 53 if (l == r) return; 54 int mid = (l + r) >> 1; 55 build(l, mid, tar << 1); 56 build(mid + 1, r, tar << 1 | 1); 57 } 58 59 void update(int kind, int l, int r, int tar) 60 { 61 if (l == t[tar].l && r == t[tar].r) 62 { 63 if (kind == 1) t[tar].ini(); 64 else if (kind == 2) t[tar].ini(), t[tar].ini1(); 65 else t[tar].full(), t[tar].full1(); 66 return; 67 } 68 pushdown(tar); 69 int mid = (t[tar].l + t[tar].r) >> 1; 70 if (r <= mid) update(kind, l, r, tar << 1); 71 else if (l > mid) update(kind, l, r, tar << 1 | 1); 72 else update(kind, l, mid, tar << 1), update(kind, mid + 1, r, tar << 1 | 1); 73 pushup(tar); 74 } 75 76 int query(int kind, int tar, int v) 77 { 78 if (kind == 1) 79 { 80 if (t[tar].mm < v) return 0; 81 if (t[tar].lm >= v) return t[tar].l; 82 if (t[tar << 1].mm >= v) return query(kind, tar << 1, v); 83 if (t[tar << 1].rm + t[tar << 1 | 1].lm >= v) return t[tar << 1].r - t[tar << 1].rm + 1; 84 return query(kind, tar << 1 | 1, v); 85 } 86 else 87 { 88 if (t[tar].mm1 < v) return 0; 89 if (t[tar].lm1 >= v) return t[tar].l; 90 if (t[tar << 1].mm1 >= v) return query(kind, tar << 1, v); 91 if (t[tar << 1].rm1 + t[tar << 1 | 1].lm1 >= v) return t[tar << 1].r - t[tar << 1].rm1 + 1; 92 return query(kind, tar << 1 | 1, v); 93 } 94 } 95 96 int main() 97 { 98 int T; cin >> T; 99 char dir[100]; 100 int x, y; 101 int cases = 0; 102 103 while (T--) 104 { 105 scanf("%d%d", &n, &m); 106 build(1, n, 1); 107 printf("Case %d:\n", ++cases); 108 while (m--) 109 { 110 scanf("%s", dir); 111 if (dir[0] == 'D') 112 { 113 scanf("%d", &x); 114 int l = query(1, 1, x); 115 //cout << l << endl; 116 if(!l) printf("fly with yourself\n"); 117 else 118 { 119 printf("%d,let's fly\n", l); 120 update(1, l, l + x - 1, 1); //kind, l, r, tar 121 } 122 } 123 else if (dir[0] == 'N') 124 { 125 scanf("%d", &x); 126 int l = query(1, 1, x); 127 //cout << l << endl; 128 if (l != 0) 129 { 130 printf("%d,don't put my gezi\n", l); 131 update(2, l, l + x - 1, 1); 132 } 133 else 134 { 135 l = query(2, 1, x); 136 if (l != 0) 137 { 138 printf("%d,don't put my gezi\n", l); 139 update(2, l, l + x - 1, 1); 140 } 141 else printf("wait for me\n"); 142 } 143 } 144 else 145 { 146 scanf("%d%d", &x, &y); 147 printf("I am the hope of chinese chengxuyuan!!\n"); 148 update(3, x, y, 1); 149 } 150 } 151 } 152 }
