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Posted on 2008-01-09 14:11 oyjpart 閱讀(7783) 評(píng)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.
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| 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)
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| Output |
For each query “Q x”, print the current number of stones in basket x.
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| Sample Input |
5 8
C 5 8
Q 5
Q 4
U 5 3
C 2 6
Q 2
Q 1
U 2 8
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| Sample Output |
2
1
4
3
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| Problem Source |
birdman
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上次比賽沒有做..補(bǔ)做一個(gè)..挺好的題..重寫了點(diǎn)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(shí)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(shí)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(diǎn)的覆蓋次數(shù)
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
good pro
but cant follow you
如果我們把這個(gè)環(huán)放成直線（準(zhǔn)確的說是一個(gè)區(qū)間）來看的話，放入某一個(gè)籃子并且按照順時(shí)針旋轉(zhuǎn)一直放num，相當(dāng)于在這個(gè)區(qū)間插入很多條線段。而進(jìn)一步說，我們可以考慮只有3中線段，比如
區(qū)間是[0,4] 從3開始插入長(zhǎng)度為11的線段 則可以分成
[3,4]
[0,4] * 2
[0,0]
而逆時(shí)針的情況很好處理,如果你現(xiàn)算出最后停在哪個(gè)點(diǎn)上,換一下起始點(diǎn)和終點(diǎn)就是順時(shí)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(tǒng)計(jì)詢問中的點(diǎn)上有多少線段覆蓋就可以了.
要進(jìn)行點(diǎn)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(jié)點(diǎn)向上到根節(jié)點(diǎn),去疊加覆蓋數(shù)就可以了.
呵呵~~
int get(int x) { // 獲得點(diǎn)的覆蓋次數(shù)
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(gè)錯(cuò)誤，你的代碼對(duì)這組數(shù)據(jù)通不過：
5 3
C 1 5
Q 1
Q 5
應(yīng)改為：
int get(int x) { // 獲得點(diǎn)的覆蓋次數(shù)
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}
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