题意:平面有k个障碍点。从(0,0)出发,第一次走1个单位,……,第n次走n个单位,恰好回到(0,0),每次必须转弯90°,图形可以自交,但不能经过障碍点。按字典序输出所有移动序列,并输出序列总数。
分析:
1、障碍点可能在出发点。
2、注意拐点不能重复!!!
3、按字典序输出。
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) typedef long long ll; typedef unsigned long long llu; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const ll LL_INF = 0x3f3f3f3f3f3f3f3f; const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {1, 0, 0, -1}; const int dc[] = {0, 1, -1, 0}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const double eps = 1e-8; const int MAXN = 500 + 10; const int MAXT = 10000 + 10; using namespace std; int vis[MAXN][MAXN]; int mark[MAXN][MAXN]; int ans[30]; int n; int cnt; map<int, char> mp; void init(){ string s = "ensw"; for(int i = 0; i < 4; ++i){ mp[i] = s[i]; } } bool judge(int x, int y, int tx, int ty, int length){ if((vis[tx][ty] || mark[tx][ty]) && !(tx == 255 && ty == 255 && length == n)) return false;//重复经过某个拐点,但此点不是走了n步后到达的原点 if(x != tx){ if(x > tx) swap(x, tx); for(int i = x + 1; i < tx; ++i){ if(vis[i][y]){ return false; } } return true; } if(y != ty){ if(y > ty) swap(y, ty); for(int i = y + 1; i < ty; ++i){ if(vis[x][i]){ return false; } } return true; } } void dfs(int x, int y, int length){ if(length == n + 1){ if(x == 255 && y == 255){ ++cnt; for(int i = 1; i <= n; ++i){ printf("%c", mp[ans[i]]); } printf("\n"); } } else{ //必须90°转弯,所以只能向两个方向走 int dir[3]; if(ans[length - 1] == 0){//东 dir[0] = 1;//北 dir[1] = 2;//南 } else if(ans[length - 1] == 1){//北 dir[0] = 0; dir[1] = 3; } else if(ans[length - 1] == 2){//南 dir[0] = 0; dir[1] = 3; } else if(ans[length - 1] == 3){//西 dir[0] = 1; dir[1] = 2; } for(int i = 0; i < 2; ++i){ int tx = x + dr[dir[i]] * length; int ty = y + dc[dir[i]] * length; if(judge(x, y, tx, ty, length)){ ans[length] = dir[i]; ++mark[tx][ty]; dfs(tx, ty, length + 1); --mark[tx][ty]; } } } } int main(){ init(); int T; scanf("%d", &T); while(T--){ memset(vis, 0, sizeof vis); memset(ans, 0, sizeof ans); memset(mark, 0, sizeof mark); cnt = 0; int k; scanf("%d%d", &n, &k); while(k--){ int x, y; scanf("%d%d", &x, &y); vis[x + 255][y + 255] = 1; } if(vis[255][255]){ printf("Found 0 golygon(s).\n\n"); continue; } int x = 255;//下标不能为负,所以所有坐标加255,原点在(255,255)。 int y = 255; int length = 1; mark[x][y] = 1; for(int i = 0; i < 4; ++i){//东,北,南,西 int tx = x + dr[i] * length; int ty = y + dc[i] * length; if(judge(x, y, tx, ty, length)){//向此方向走length步无障碍物 ans[length] = i; ++mark[tx][ty]; dfs(tx, ty, length + 1); --mark[tx][ty]; } } printf("Found %d golygon(s).\n\n", cnt); } return 0; }