【CodeVS 3290】【NOIP 2013】华容道
http://codevs.cn/problem/3290/
据说2013年的noip非常难,但Purpleslz学长还是AK了。能A掉这道题真心orz。
设状态(i,j,k)表示目标棋子在(i,j)这个位置,空格在紧贴着目标棋子的k方向,0≤k<4。
因为目标棋子要移动,空格肯定在它旁边。往空格的方向走一步,空格便出现在它另一边。对于这两个状态连边,边权为1。
为了使目标棋子向某一方向移动,需要目标棋子不动,空格从紧贴着目标棋子的某一方向移动到紧贴着目标棋子的另一个方向。对于固定目标棋子位置但空格对于目标棋子的方向不同的状态之间互相连边,边权需要bfs求得。
对于每个询问,给出初始空格的位置,bfs出初始的空格移动到目标棋子旁边四个位置的最短距离,并连边, 边权为最短距离。
最后跑spfa求出到达目标棋子到达终点需要走的最短路。
时间复杂度O(nm),因为边数是nmk级别的,而且k是个常数,所以把k忽略掉233 _(:з」∠)_k都快比nm大了QwQ
话说spfa的复杂度真的是O(E)的吗(/"≡ _ ≡)/~┴┴
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 | #include<queue> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 33; const int dx[4] = {-1, 0, 1, 0}; const int dy[4] = {0, 1, 0, -1}; int in() { int k = 0, fh = 1; char c = getchar (); for (; c < '0' || c > '9' ; c = getchar ()) if (c == '-' ) fh = -1; for (; c >= '0' && c <= '9' ; c = getchar ()) k = (k << 3) + (k << 1) + c - '0' ; return k * fh; } struct node { int nxt, to, w;} E[20003]; int Move[33][33][4][4], point[4003], cnt = 0, a[33][33], n, m, qq; int number[33][33][4], tot = 0; void ins( int u, int v, int w) {E[++cnt] = (node) {point[u], v, w}; point[u] = cnt;} bool can( int x, int y) { return ((x >= 1) && (x <= n) && (y >= 1) && (y <= m) && (a[x][y] == 1)); } struct Point { int x, y, d; Point( int _x = 0, int _y = 0, int _d = 0) : x(_x), y(_y), d(_d) {} bool operator == ( const Point &A) const { return x == A.x && y == A.y; } } q[1003]; int cross( int x, int y, int h, int hh) { Point s, t; s = Point(x + dx[h], y + dy[h], 0); t = Point(x + dx[hh], y + dy[hh], 0); a[x][y] = 0; int head = 0, tail = 1; q[1] = s; a[s.x][s.y] = 1; Point u, v; while (head != tail) { u = q[++head]; if (u == t) break ; for ( int d = 0; d < 4; ++d) if (can(u.x + dx[d], u.y + dy[d])) { v = Point(u.x + dx[d], u.y + dy[d], u.d + 1); q[++tail] = v; a[v.x][v.y] = 0; } } for ( int i = 1; i <= tail; ++i) a[q[i].x][q[i].y] = 1; a[x][y] = 1; if (u == t) return u.d; else return 0x7fffffff; } void dealwith( int x, int y) { int ed; for ( int d = 0; d < 4; ++d) if (can(x + dx[d], y + dy[d])) for ( int dd = d + 1; dd < 4; ++dd) if (can(x + dx[dd], y + dy[dd])) { ed = Move[x][y][d][dd] = Move[x][y][dd][d] = cross(x, y, d, dd); if (ed != 0x7fffffff) { ins(number[x][y][d], number[x][y][dd], ed); ins(number[x][y][dd], number[x][y][d], ed); } } } int dis(Point g, Point s, Point t) { int head = 0, tail = 1; s.d = 0; q[1] = s; a[s.x][s.y] = 0; a[g.x][g.y] = 0; Point u, v; while (head != tail) { u = q[++head]; if (u == t) break ; for ( int d = 0; d < 4; ++d) if (can(u.x + dx[d], u.y + dy[d])) { v = Point(u.x + dx[d], u.y + dy[d], u.d + 1); q[++tail] = v; a[v.x][v.y] = 0; } } a[g.x][g.y] = 1; for ( int i = 1; i <= tail; ++i) a[q[i].x][q[i].y] = 1; if (u == t) return u.d; else return 0x7fffffff; } queue < int > Q; int dist[4003], inq[4003]; void spfa( int s) { memset (dist, 127, sizeof ( int ) * (tot + 1)); Q.push(s); dist[s] = 0; inq[s] = true ; int u; while (!Q.empty()) { u = Q.front(); Q.pop(); inq[u] = false ; for ( int i = point[u]; i; i = E[i].nxt) if (dist[u] + E[i].w < dist[E[i].to]) { dist[E[i].to] = dist[u] + E[i].w; if (!inq[E[i].to]) { Q.push(E[i].to); inq[E[i].to] = true ; } } } } int main() { n = in(); m = in(); qq = in(); for ( int i = 1; i <= n; ++i) for ( int j = 1; j <= m; ++j) a[i][j] = in(); for ( int i = 1; i <= n; ++i) for ( int j = 1; j <= m; ++j) if (a[i][j] == 1) for ( int d = 0; d < 4; ++d) if (can(i + dx[d], j + dy[d])) number[i][j][d] = ++tot; for ( int i = 1; i <= n; ++i) for ( int j = 1; j <= m; ++j) if (a[i][j] == 1) dealwith(i, j); for ( int i = 1; i <= n; ++i) for ( int j = 1; j <= m; ++j) if (a[i][j] == 1) for ( int d = 0; d < 4; ++d) if (can(i + dx[d], j + dy[d])) ins(number[i][j][d], number[i + dx[d]][j + dy[d]][(d + 2) % 4], 1); ++tot; Point e, s, t; int now = cnt, ans; while (qq--) { e.x = in(); e.y = in(); s.x = in(); s.y = in(); t.x = in(); t.y = in(); if (s == t) { puts ( "0" ); continue ;} cnt = now; point[tot] = 0; for ( int d = 0; d < 4; ++d) if (can(s.x + dx[d], s.y + dy[d])) ins(tot, number[s.x][s.y][d], dis(s, e, Point(s.x + dx[d], s.y + dy[d], 0))); spfa(tot); ans = 0x7fffffff; for ( int d = 0; d < 4; ++d) ans = min(ans, dist[number[t.x][t.y][d]]); printf ( "%d\n" , ans == 2139062143 ? -1 : ans); } return 0; } |
终于A掉了。注意特判起点和终点相同
NOI 2017 Bless All
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