POJ 3009 Curling 2.0

 ?1 #pragma GCC optimize("Ofast") ?2 #include <bits/stdc++.h> ?3 using namespace std; ?4 ???5 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0); ?6 #define Rep(i,n) for (int i = 0; i < (n); ++i) ?7 #define For(i,s,t) for (int i = (s); i <= (t); ++i) ?8 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) ?9 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 10 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 11 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 12 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 13 ??14 #define pr(x) cout << #x << " = " << x << " ?" 15 #define prln(x) cout << #x << " = " << x << endl 16 ??17 #define LOWBIT(x) ((x)&(-x)) 18 ??19 #define ALL(x) x.begin(),x.end() 20 #define INS(x) inserter(x,x.begin()) 21 ??22 #define ms0(a) memset(a,0,sizeof(a)) 23 #define msI(a) memset(a,inf,sizeof(a)) 24 #define msM(a) memset(a,-1,sizeof(a)) 25 ?26 #define MP make_pair 27 #define PB push_back 28 #define ft first 29 #define sd second 30 ??31 template<typename T1, typename T2> 32 istream &operator>>(istream &in, pair<T1, T2> &p) { 33 ????in >> p.first >> p.second; 34 ????return in; 35 } 36 ??37 template<typename T> 38 istream &operator>>(istream &in, vector<T> &v) { 39 ????for (auto &x: v) 40 ????????in >> x; 41 ????return in; 42 } 43 ??44 template<typename T1, typename T2> 45 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 46 ????out << "[" << p.first << ", " << p.second << "]" << "\n"; 47 ????return out; 48 } 49 ??50 typedef long long LL; 51 typedef unsigned long long uLL; 52 typedef pair< double, double > PDD; 53 typedef pair< int, int > PII; 54 typedef pair< LL, LL > PLL; 55 typedef set< int > SI; 56 typedef vector< int > VI; 57 typedef map< int, int > MII; 58 typedef vector< LL > VL; 59 typedef vector< VL > VVL; 60 const double EPS = 1e-10; 61 const int inf = 1e9 + 9; 62 const LL mod = 1e9 + 7; 63 const int maxN = 1e5 + 7; 64 const LL ONE = 1; 65 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 66 const LL oddBits = 0x5555555555555555; 67 ?68 int n, m, ans; 69 int M[23][23]; 70 PII S, G; 71 ?72 // 从 x 点到 y 点是否经过目标点 ?73 bool isPassed(PII x, PII y) { 74 ????if(x.ft == y.ft) 75 ????????return x.ft == G.ft && min(x.sd, y.sd) <= G.sd && G.sd <= max(x.sd, y.sd); 76 ????else ?77 ????????return x.sd == G.sd && min(x.ft, y.ft) <= G.ft && G.ft <= max(x.ft, y.ft); 78 } ?79 ?80 // 找到 x 最左边的可到达点 ?81 PII leftMost(PII x) { 82 ????while(x.sd > 0 && M[x.ft][x.sd - 1] != 1) --x.sd; 83 ????return x; 84 } 85 ?86 // 找到 x 最上边的可到达点 ?87 PII upMost(PII x) { 88 ????while(x.ft > 0 && M[x.ft - 1][x.sd] != 1) --x.ft; 89 ????return x; 90 } 91 ?92 // 找到 x 最右边的可到达点 ?93 PII rightMost(PII x) { 94 ????while(x.sd < m - 1 && M[x.ft][x.sd + 1] != 1) ++x.sd; 95 ????return x; 96 } 97 ?98 // 找到 x 最下边的可到达点 ?99 PII downMost(PII x) {100 ????while(x.ft < n - 1 && M[x.ft + 1][x.sd] != 1) ++x.ft;101 ????return x;102 }103 104 // 以数组方式存储函数 105 PII (*func[4])(PII) = {leftMost, upMost, rightMost, downMost};106 int dx[] = {0, -1, 0, 1};107 int dy[] = {-1, 0, 1, 0};108 109 void dfs(PII x, int cnt) {110 ????if(cnt > 10) return;111 ????if(x == G) ans = min(ans, cnt);112 ????113 ????Rep(i, 4) {114 ????????PII tmp = func[i](x);115 ????????// 是否经过G 116 ????????if(isPassed(x, tmp)) ans = min(ans, cnt + 1);117 ????????// 如果能移动到不同地方且不再边界上 118 ????????if(x != tmp && tmp.ft + dx[i] >= 0 && tmp.ft + dx[i] < n && tmp.sd + dy[i] >= 0 && tmp.sd + dy[i] < m) {119 ????????????M[tmp.ft + dx[i]][tmp.sd + dy[i]] = 0;120 ????????????dfs(tmp, cnt + 1);121 ????????????M[tmp.ft + dx[i]][tmp.sd + dy[i]] = 1;122 ????????}123 ????}124 }125 126 int main(){127 ????INIT();128 ????while(cin >> m >> n) {129 ????????if(m == 0) break;130 ????????ans = inf;131 ????????Rep(i, n) {132 ????????????Rep(j, m) {133 ????????????????cin >> M[i][j];134 ????????????????if(M[i][j] == 2) S = MP(i, j);135 ????????????????if(M[i][j] == 3) G = MP(i, j);136 ????????????}137 ????????}138 ????????dfs(S, 0);139 ????????if(ans > 10) ans = -1;140 ????????cout << ans << endl;141 ????}142 ????return 0;143 }