USACO Wormholes 解题日志

表示PPCA做题还是挺快乐的。虽然还不知道那个大作业到底是什么题。

先贴上这道题的题目：

Wormholes

Farmer John's hobby of conducting high-energy physics experiments on weekends has backfired, causing N wormholes (2 <= N <= 12, N even) to materialize on his farm, each located at a distinct point on the 2D map of his farm (the x,y coordinates are both integers).

According to his calculations, Farmer John knows that his wormholes will form N/2 connected pairs. For example, if wormholes A and B are connected as a pair, then any object entering wormhole A will exit wormhole B moving in the same direction, and any object entering wormhole B will similarly exit from wormhole A moving in the same direction. This can have rather unpleasant consequences.

For example, suppose there are two paired wormholes A at (1,1) and B at (3,1), and that Bessie the cow starts from position (2,1) moving in the +x direction. Bessie will enter wormhole B [at (3,1)], exit from A [at (1,1)], then enter B again, and so on, getting trapped in an infinite cycle!

   | . . . .
   | A > B .      Bessie will travel to B then
   + . . . .      A then across to B again

Farmer John knows the exact location of each wormhole on his farm. He knows that Bessie the cow always walks in the +x direction, although he does not remember where Bessie is currently located.

Please help Farmer John count the number of distinct pairings of the wormholes such that Bessie could possibly get trapped in an infinite cycle if she starts from an unlucky position. FJ doesn't know which wormhole pairs with any other wormhole, so find all the possibilities.

PROGRAM NAME: wormhole

INPUT FORMAT:

Line 1:	The number of wormholes, N.
Lines 2..1+N:	Each line contains two space-separated integers describing the (x,y) coordinates of a single wormhole. Each coordinate is in the range 0..1,000,000,000.

SAMPLE INPUT (file wormhole.in):

4
0 0
1 0
1 1
0 1

INPUT DETAILS:

There are 4 wormholes, forming the corners of a square.

OUTPUT FORMAT:

Line 1:

The number of distinct pairings of wormholes such that Bessie could conceivably get stuck in a cycle walking from some starting point in the +x direction.

SAMPLE OUTPUT (file wormhole.out):

2

OUTPUT DETAILS:

If we number the wormholes 1..4 as we read them from the input, then if wormhole 1 pairs with wormhole 2 and wormhole 3 pairs with wormhole 4, Bessie can get stuck if she starts anywhere between (0,0) and (1,0) or between (0,1) and (1,1).

   | . . . .
   4 3 . . .      Bessie will travel to B then
   1-2-.-.-.      A then across to B again

Similarly, with the same starting points, Bessie can get stuck in a cycle if the pairings are 1-3 and 2-4 (if Bessie enters WH#3 and comes out at WH#1, she then walks to WH#2 which transports here to WH#4 which directs her towards WH#3 again for a cycle).

Only the pairings 1-4 and 2-3 allow Bessie to walk in the +x direction from any point in the 2D plane with no danger of cycling. 

一开始还以为要用什么数据结构，把这道题想复杂了。后来Google了一下（QAQ好难过），发现网上有很多都是官方爸爸的做法。

我们采用直接开数组记录右邻居的方法：对于一个洞A，用nextRight[A]记下和它右边的和它纵坐标相等的洞中最靠近它的洞的下标。然后如果要判断有没有环，依次走下去就好了。具体实现在函数exiCycle()中，就不用多说。

然后注意的一点是，我们的pair[i]，pair[j]中的i，j不是横坐标值，而是横坐标对应的输入时的下标值。一开始做的时候不小心搞混了，查了好久QAQ。

下面是代码：

#include<fstream>
#define maxn 15
using namespace std;

ifstream fin("wormhole.in");
ofstream fout("wormhole.out");

int n, x[maxn], y[maxn];
//int total;
int nextRight[maxn];
int pairs[maxn];

bool exiCycle()
{
        for (int i = 1; i <= n; ++i){
                int pos = i;
                for (int j = 1; j <= n; ++j)
                        pos = nextRight[pairs[pos]];

                if (pos != 0) return 1;
        }
        return 0;
}

int solve()
{
        int i = 1;
        int total = 0;
        for (i = 1; i <= n; ++i)
                if (pairs[i] == 0) break;

        if (i > n){
                if (exiCycle()) return 1;
                else return 0;
        }

        for (int j = i + 1; j <= n; ++j){
                if (pairs[j] == 0){
                        pairs[i] = j;
                        pairs[j] = i;
                        total += solve();
                        pairs[i] = pairs[j] = 0;
                }
        }

        return total;
}

int main()
{
        fin >> n;
        for (int i = 1; i <= n; ++i) fin >> x[i] >> y[i];

        for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j){
                if (y[i] == y[j] && x[i] < x[j]){
                        if (nextRight[i] == 0 || x[nextRight[i]] - x[i] > x[j] - x[i])
                                nextRight[i] = j;
                }
        }

        fout << solve() << endl;

        fin.close();
        fout.close();

        return 0;
}