2014
01-26

# Human Knot

A classic ice-breaking exercise is for a group of n people to form a circle and then arbitrarily join hands with one another. This forms a "human knot" since the players’ arms are most likely intertwined. The goal is then to unwind the knot to form a circle of players with no arms crossed.

We now adapt this game to a more general and more abstract setting where the physical constraints of the problem are gone. Suppose we represent the initial knot with a 2-regular graph inscribed in a circle (i.e., we have a graph with n vertices with exactly two edges incident on each vertex). Initially, some edges may cross other edges and this is undesirable. This is the "knot" we wish to unwind.

A "move" involves moving any vertex to a new position on the circle, keeping its edges intact. Our goal is to make the fewest possible moves such that we obtain one n-sided polygon with no edge-crossings remaining.

For example, here is a knot on 4 vertices inscribed in a circle, but two edges cross each other. By moving vertex 4 down to the position between 2 and 3, a graph without edge-crossings emerges. This was achieved in a single move, which is clearly optimal in this case.

When n is larger, things may not be quite as clear. Below we see a knot on 6 vertices. We might consider moving vertex 4 between 5 and 6, then vertex 5 between 1 and 2, and finally vertex 6 between 3 and 4; this unwinds the knot in 3 moves.

But clearly we can unwind the same knot in only two moves:

The input consists of a number of cases. Each case starts with a line containing the integer n (3 <= n <= 500), giving the number of vertices of the graph. The vertices are labelled clockwise from 1 to n. Each of the next n lines gives a pair of neighbors, where line i (1 <= i <= n) specifies the two vertices adjacent to vertex i. The input is terminated by n = 0.

The input consists of a number of cases. Each case starts with a line containing the integer n (3 <= n <= 500), giving the number of vertices of the graph. The vertices are labelled clockwise from 1 to n. Each of the next n lines gives a pair of neighbors, where line i (1 <= i <= n) specifies the two vertices adjacent to vertex i. The input is terminated by n = 0.

6
4 5
3 5
2 6
1 6
1 2
3 4
6
2 6
1 4
5 6
2 5
3 4
1 3
0

Knot solvable.
2
Knot solvable.
1

1. 您没有考虑 树的根节点是负数的情况， 若树的根节点是个很大的负数，那么就要考虑过不过另外一边子树了

2. 有一点问题。。后面动态规划的程序中
int dp[n+1][W+1];
会报错 提示表达式必须含有常量值。该怎么修改呢。。