|
|
Posted on 2008-01-09 14:11 oyjpart 閱讀(7747) 評論(8) 編輯 收藏 引用 所屬分類: ACM/ICPC或其他比賽
| Number of stones |
| Time Limit: 3000ms, Special Time Limit:7500ms, Memory Limit:32768KB |
| Total submit users: 13, Accepted users: 1 |
| Problem 11107 : No special judgement |
| Problem description |
There are N baskets rounded in a circle, and numbered as 1、2、3、…….、N-1、N, in clockwise. At the beginning, all of the baskets are empty. Some workers go to the moutain to collect stones. When they are back,they put their stones to some baskets. The workers have a habit, Once a worker come back, he choose a baskets, and choose a direction(clockwise or anticlockwise), he put one stone to this basket and move to the next basket according to the direction he has chosen, he continues doing this until all of the stones they have collected have been put down.
Sometimes the workers ask you how many stone it is in the basket, as there are too many baskets, You are to wirte a program to calculate this.
|
| Input |
The input test file will contain multiple cases. Each test case:
In the first line of the input contain two integers N,M(3 <= N <= 100000, 0 <= M <= 300000). Following M lines,each line represents an event. There are only three kinds of events: Q, C, U. And the format is:
“Q x”, query the number of stones in the basket x.
“C x num”, a worker comes back and the number of the stones he has picked up is num, he puts down stones from the basket x in clockwise.
“U x num”, a worker comes back and the number of the stone he has picked up is num, he puts down stones from the basket x in anticlockwise.
(x, num are both integers, 1 <= x <= N, 1 <= num <= 10000)
|
| Output |
For each query “Q x”, print the current number of stones in basket x.
|
| Sample Input |
5 8
C 5 8
Q 5
Q 4
U 5 3
C 2 6
Q 2
Q 1
U 2 8
|
| Sample Output |
2
1
4
3
|
| Problem Source |
birdman
|
上次比賽沒有做..補做一個..挺好的題..重寫了點樹模板
1  /**//*
2 * 主程序要作的事情
3 * 1.確定N ：必須是2^n，可以取實際n的上界
4 * 2.build(left, right);
5 *
6 */
7
8#include <cstdio>
9#include <cstring>
10
11const int N = 131072; //必須是2^n，可以取實際n的上界
12
13int upperbound;
14
15struct Node  {
16 int i, j, c, m; //left, right
17} T[N*2];
18
19void bd(int d, int left, int right) {
20 T[d].i = left, T[d].j = right, T[d].c = 0;
21  if(right > left) {
22 bd(d*2+1, left, T[d].m = (left+right)>>1);
23 bd(d*2+2, T[d].m+1, right);
24 }
25}
26
27void build(int left, int right) {
28 upperbound = 1;
29 while(upperbound < right-left+1) upperbound <<= 1;
30 bd(0, left, left + upperbound-1);
31}
32
33void add(int d, int left, int right, int c) {
34 if(left <= T[d].i && right >= T[d].j) {
35 T[d].c += c;
36 }
37 else {
38 if(left <= T[d].m) add(d*2+1, left, right, c);
39 if(right > T[d].m) add(d*2+2, left, right, c);
40 }
41}
42
43int get(int x) { // 獲得點的覆蓋次數
44 int idx = upperbound+x-1, sum = 0;
45 do {
46 sum += T[idx].c;
47 idx = (idx-1)>>1;
48 } while(idx != 0);
49 return sum;
50}
51
52int n, m;
53
54void Add(int x, int num) {
55 int laps = (num-(n-x))/n;
56 if(laps > 0) {
57 add(0, 0, n-1, laps);
58 }
59 num -= laps*n;
60 if(n-x >= num) {
61 add(0, x, x+num-1, 1);
62 }
63 else {
64 add(0, x, n-1, 1);
65 add(0, 0, (x+num-1)%n, 1);
66 }
67}
68
69int main() {
70 while(scanf("%d %d", &n, &m) != EOF) {
71 build(0, n-1);
72 while(m--) {
73 char cmd;
74 int x, num;
75 scanf(" %c", &cmd);
76 if(cmd == 'Q') {
77 scanf("%d", &x);
78 --x;
79 printf("%d\n", get(x));
80 }
81 else if(cmd == 'C') {
82 scanf("%d %d", &x, &num);
83 --x;
84 Add(x, num);
85 }
86 else if(cmd == 'U') {
87 scanf("%d %d", &x, &num);
88 --x;
89 int y = (x-num+1) % n;
90 if(y < 0) y += n;
91 Add(y, num);
92 }
93 }
94 }
95
96 return 0;
97}
Feedback
# re: HOJ 11107 回復 更多評論
2008-05-24 21:25 by
good pro
but cant follow you
# re: HOJ 11107 回復 更多評論
2008-05-25 20:31 by
如果我們把這個環放成直線（準確的說是一個區間）來看的話，放入某一個籃子并且按照順時針旋轉一直放num，相當于在這個區間插入很多條線段。而進一步說，我們可以考慮只有3中線段，比如
區間是[0,4] 從3開始插入長度為11的線段 則可以分成
[3,4]
[0,4] * 2
[0,0]
而逆時針的情況很好處理,如果你現算出最后停在哪個點上,換一下起始點和終點就是順時針了.
最后是線段樹了,我們把所有的線段都分別插入.最后統計詢問中的點上有多少線段覆蓋就可以了.
要進行點的線段覆蓋查詢,有很多種做法,我覺得比較好的就是從葉節點向上到根節點,去疊加覆蓋數就可以了.
呵呵~~
# re: HOJ 11107 回復 更多評論
2008-06-03 14:01 by
建樹可以非遞歸話吧
# re: HOJ 11107 回復 更多評論
2008-10-13 10:57 by
謝謝大牛了，我搞了半天終于弄懂了什么原理哈
# re: HOJ 11107 回復 更多評論
2008-10-13 14:14 by
int get(int x) { // 獲得點的覆蓋次數
44 int idx = upperbound+x-1, sum = 0;
45 do {
46 sum += T[idx].c;
47 idx = (idx-1)>>1;
48 } while(idx != 0);
49 return sum;
50}
貌似這里有個錯誤，你的代碼對這組數據通不過：
5 3
C 1 5
Q 1
Q 5
應改為：
int get(int x) { // 獲得點的覆蓋次數
44 int idx = upperbound+x-1, sum = 0;
45 do {
46 sum += T[idx].c;
47 idx -= 1;
if(idx != -1) idx >>= 1;
48 } while(idx >= 0);
49 return sum;
50}
# re: HOJ 11107 回復 更多評論
2008-10-16 02:30 by
Thx!~
# re: HOJ 11107 回復 更多評論
2009-03-23 20:41 by
絕對大牛
# re: HOJ 11107 回復 更多評論
2009-03-26 00:27 by
額
|