UVA225Golygons 黄金图形

题意：平面上有k个障碍点。从（0，0）出发，第一次走一个单位，第二次走2个单位，····第n次走n个单位，恰好回到（0，0）点。要求只能沿着东南西北方向走，且每次必须转弯90度（不能沿一个方向继续走，也不能后退）。走出的图形可以自交但不能经过障碍点，已经访问过的点不能再访问。输入n、k（1n20，0k50)和所有障碍点的坐标，所有满足条件的移动序列（用news，表示北、东、西、南），按照字典序从小到大输出，最后输出移动序列总数。

分析：本题是一道典型的回溯题，需剪枝，当当前的坐标到原点的x和y的距离和大于可以移动的步数，剪枝。按照四个方向搜索，每次搜索都判断是否经过障碍物。访问过的点不能再访问。

#include<cstdio>
#include<cstring>
#include<cctype>
#include<queue>
#include<iostream>
#include<vector>
#include<list>
#include<set>
using namespace std;
const int maxn = 300;
int originx = 150, originy = 150;//坐标原点
int G[maxn][maxn]; int n;
char route[maxn]; int rous;
int dy[] = { 0,1,-1,0 };
int dx[] = { 1,0,0,-1 };
char dir[] = { 'e','n','s','w' };
int vis[maxn][maxn];
void solve(int cur,int x,int y,int direction) {
	if (cur == n + 1) {
		if (x == originx && y == originy) {
			for (int i = 1; i <= n; i++)cout << route[i];
			cout << endl;
			rous++;
		}
		return;
	}
	if (cur > n + 1)return;
	for (int i = 0; i < 4; i++) {
		if (i == direction||i==3-direction)continue;
		int ddx = x + dx[i] * cur;
		int ddy = y + dy[i] * cur;
		if (vis[ddx][ddy])continue;
		if((n - cur)*(n + cur + 1) / 2<(abs(ddx-originx)+abs(ddy-originy)))continue;
		int ok = 0;
		int a1, a2,a;
		if (ddx == x) { a = x; a1 = y; a2 = ddy; }
		else { a = y; a1 = x; a2 = ddx; }
		if (a1 > a2) swap(a1, a2);
		for (int j = a1; j <= a2; j++) {
			if (ddx == x)if (G[a][j]) {
				ok = 1; break;
			}
			if (ddy == y)if (G[j][ddy]) {
				ok = 1; break;
			}
		}
		if (!ok) {
			route[cur] = dir[i];
			vis[ddx][ddy] = 1;
			solve(cur + 1, ddx, ddy, i);
			vis[ddx][ddy] = 0;
		}
	}
}
int main() {
	int kase;
	cin >> kase;
	while (kase-- > 0) {
		memset(G, 0, sizeof(G));
		memset(vis, 0, sizeof(vis));
		rous = 0;
		cin >> n;
		int bar,barx,bary;
		cin >> bar;
		for (int i = 0; i < bar; i++) {
			cin >> barx >> bary;
			G[barx + originx][bary + originy] = 1;
		}
		solve(1, originx, originy, -1);
		cout << "Found " << rous << " golygon(s)." << endl;
		cout << endl;
	}
	//system("pause");
	return 0;
}