POJ 1988 Cube Stacking（带权并查集）
Cube Stacking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 23678 | Accepted: 8299 | |
| 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
题目链接：POJ 1988
以前看了很久的题解但是完全看不懂，也是最近会了食物链之后才做的。
既然食物链可以用数值代表关系，那这题只要把取模去掉就可以用数值node[x].up代表自己与祖节点之间的砖块数（不包括本身），然后还有一个问题就是还要知道自己所在砖块堆的总个数node[fx].total（fx是此砖块堆的祖节点即代表）。
那统计up用食物链的思想比较简单就可以做到，关键就是total怎么算，可以设想（画图更好理解），把砖a所在堆放到砖b所在堆，那node[fa].total肯定增加了node[fb].total个，然后node[fa].up显然不变（实际上祖节点就是堆的最上面那块砖的号码），node[fb].total就不要动了，因为只有祖节点才能具有记录total的功能，fb此时已沦为fa下面的一个子节点，node[fb].up还是可以动的，即增加了node[fa].total个。然后这题就差不多解决了，最后输出的时候用本堆砖块数total-本砖块上面的砖块数up-1，减1是因为题目要求只算压在自己这块砖下面的，自己当然不算
代码：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=300010;
struct info
{
int pre;
int up;
int total;
};
info node[N];
void init(int n)
{
for (int i=0; i<=n; ++i)
{
node[i].pre=i;
node[i].up=0;
node[i].total=1;
}
}
int find(int n)
{
if(node[n].pre==n)
return n;
int tpre=node[n].pre;
node[n].pre=find(node[n].pre);
node[n].up=node[n].up+node[tpre].up;
return node[n].pre;
}
void joint(int a,int b)
{
int fa=find(a),fb=find(b);
if(fa!=fb)
{
node[fb].pre=fa;
node[fb].up+=node[fa].total;
node[fa].total+=node[fb].total;
}
}
int main(void)
{
int n,i,j,a,b,x;
char ops[3];
while (~scanf("%d",&n))
{
init(n);
for (i=0; i<n; ++i)
{
scanf("%s",ops);
if(ops[0]=='M')
{
scanf("%d%d",&a,&b);
joint(a,b);
}
else
{
scanf("%d",&x);
int fx=find(x);
printf("%d\n",node[fx].total-node[x].up-1);
}
}
}
return 0;
}
