POJ 1988 并查集 妙用deep数组
2016-08-05 20:45
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[align=center]Cube Stacking[/align]
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.
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.
Output
Print the output from each of the count operations in the same order as the input file.
Sample Input
Sample Output
题意:搬箱子和架箱子,如果箱子上面有箱子,则再不打乱顺序的条件下把整体搬过去
先普及下基础:并查集就是在路径压缩的条件下找根,每次在函数返回的时候,顺带把路上遇到的人的BOSS改为最后找到的祖宗编号。这样可以提高今后找到最高领导人(也就是树的祖先)的速度。然后再merge中判断是不是在一个连通分量,若不是,则连通,把一个连通图的根赋给另一个连通图,靠这样的方式去合并一个连通分量。
看了网上大牛的思路做出来的,在并查集的基础上加一个deep数组,用来存旧根的深度,旧根的深度等于新根上一次的节点数,这里就是一个线树,最后输出的时候,查根,减去改节点的深度,再减去本身,就得到下面的箱子,很巧妙啊,整体分为了根的节点数,本身,本身的深度三部分,思路实在巧妙。
如 给出这个数据:
Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 23635 | Accepted: 8281 | |
Case Time Limit: 1000MS |
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.
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.
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
题意:搬箱子和架箱子,如果箱子上面有箱子,则再不打乱顺序的条件下把整体搬过去
先普及下基础:并查集就是在路径压缩的条件下找根,每次在函数返回的时候,顺带把路上遇到的人的BOSS改为最后找到的祖宗编号。这样可以提高今后找到最高领导人(也就是树的祖先)的速度。然后再merge中判断是不是在一个连通分量,若不是,则连通,把一个连通图的根赋给另一个连通图,靠这样的方式去合并一个连通分量。
看了网上大牛的思路做出来的,在并查集的基础上加一个deep数组,用来存旧根的深度,旧根的深度等于新根上一次的节点数,这里就是一个线树,最后输出的时候,查根,减去改节点的深度,再减去本身,就得到下面的箱子,很巧妙啊,整体分为了根的节点数,本身,本身的深度三部分,思路实在巧妙。
如 给出这个数据:
#include<cstdio> #include<iostream> #include<cstring> using namespace std; const int N =30010; int father , jiedian , deep ; int find(int x) { int temp; if(x == father[x]) return x; temp = father[x]; father[x] = find(temp); deep[x] += deep[temp]; //与根结点的距离 return father[x]; } int main() { //freopen("in.txt","r",stdin); int p; char ope; int a, b; int query; int root1, root2; scanf("%d", &p); for(int i = 1; i < N; ++i) //init { father[i] = i; jiedian[i] = 1; deep[i] = 0; } for(int i = 0; i < p; ++i) { scanf("%*c%c", &ope); if(ope == 'M') { scanf("%d%d", &a, &b); root1 = find(a); root2 = find(b); if(root1 != root2) { father[root2] = root1; deep[root2] = jiedian[root1]; //更新深度,这里的深度指的是root2箱子上箱子的个数,因为root1架在上面,离根的距离 jiedian[root1] += jiedian[root2]; //更新结点数,根下面的节点,包括根 } } else { scanf("%d", &query); printf("%d\n", jiedian[find(query)] - deep[query] - 1); //减去自己 } } return 0; }
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