Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 15951 | Accepted: 5448 | |
Case Time Limit: 1000MS |
Description
Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations:
moves and counts.
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.
Write a program that can verify the results of the game.
moves and counts.
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.
Write a program that can verify the results of the game.
Input
* Line 1: A single integer, P
* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.
Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.
* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.
Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.
Output
Print the output from each of the count operations in the same order as the input file.
Sample Input
6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4
Sample Output
1 0 2
Source
题意:给你一定数目的堆(PS:堆的数目不确定,但是不超过最大值 30,000)
初始时每个堆都有且仅有一个方块 ,其编号安装所在堆的编号编为 1 ... 30,000 。
输入一个数 P,表示有 P 个操作:
C x 计算编号为 x 的方块下面有多少个方块,输出结果。
M x y 将编号为 x 的方块所在的堆放到编号为 y 所在的堆的上面。
算法:并查集。
思路【盗版的,惭愧】:
以堆底为根【重要】(PS:因为方便用堆底的下标记录当前堆的总方块数)
p[] 记录父节点
r[] 记录当前方块下面的方块个数
sum[] 记录当前堆的方块总个数,记录堆底【根】的即可。
并查集模板:
1.找根节点:注意向下更新时 r[] 要加上父亲的 r[]
2.合并:因为是前者放在后者上面,又是应该以堆底为根,所以 p[fx] = fy;
相应要更新 r[fx] = num[fy]; // x 的根【堆底】压 y 所在堆的所有方块块
num[fy] += num[fx]; // 被压的堆的方块总数更新
G | Accepted | 484 KB | 250 ms | C++ | 1173 B | 2013-04-10 16:48:07 |
#include<cstdio> const int maxn = 30000+10; int p[maxn]; int r[maxn]; //下面的个数 int sum[maxn]; //总个数 int find(int x) { if(x == p[x]) return x; int t = p[x]; p[x] = find(p[x]); r[x] = r[x]+r[t]; //自己下面的+父亲下面的 return p[x]; } void Union(int x, int y) { int fx = find(x); int fy = find(y); if(fx == fy) return; p[fx] = fy; //以堆底为根 r[fx] = sum[fy]; //上面的堆底压了 y所在堆的所有木块 sum[fy] = sum[fy]+sum[fx]; //更新当前堆的木块个数 保存在根中 } void set() { for(int x = 1; x < maxn; x++) { p[x] = x; r[x] = 0; sum[x] = 1; } } int main() { int p; while(scanf("%d%*c", &p) != EOF) { set(); char c; int x,y; while(p--) { scanf("%c", &c); if(c == 'M') { scanf("%d%d%*c", &x, &y); Union(x, y); } else if(c == 'C') { scanf("%d%*c", &x); find(x); //可能没有更新 r[] 输出前,更新根节点 printf("%d\n", r[x]); } } } return 0; }