usaco.section1.3 wormhole(枚举)

来源：互联网发布：mac系统打不开exe文件编辑：程序博客网时间：2024/05/07 07:31

有 N 个虫洞，将这些虫洞两两分组，同组的两个虫洞可以互相到达，使得奶牛从坐标系某个位置出发，一直沿+x 方向走时会陷入无限循环

因为 N <= 12 比较小，可以枚举所有的两两连接情况，然后判断可行不可行

#include <cstdio>using namespace std;const int MAX_N=12;int N, x[MAX_N + 1], y[MAX_N + 1];int next[MAX_N + 1], p[MAX_N + 1];bool check(void){for (int i=1; i<=N; i++){int pos = i; // every startfor (int j=1; j<=N; j++) pos = next[p[pos]];if(pos != 0) return true;}return false;}int doit(void){// counting all possible casesint i, ret=0;for (i=1; i<=N; i++)  if(p[i] == 0) { i=i; break; } // finding the wormhole haven't be paired  if (i>N){ // there is not unpaired wormholeif(check())return 1;return 0;}for (int j=i+1; j<=N; j++) // pairing i with j  if(p[j] == 0){  p[j] = i; p[i] = j;  ret += doit();  p[j] = p[i] = 0;  }return ret;}int main(void){freopen("wormhole.in", "r", stdin);freopen("wormhole.out", "w", stdout);scanf("%d", &N);for (int i=1; i<=N; i++) scanf("%d%d", &x[i], &y[i]); // cinfor (int i=1; i<=N; i++)// finding next[]  for (int j=1; j<=N; j++){    if (x[i] < x[j] && y[i] == y[j]){  if (next[i] == 0 || x[j] - x[i] < x[next[i]] - x[i]) // haven't got or closernext[i] = j;}  }printf("%d\n", doit());return 0;}

0 0