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这道题的题解很多，所以就不用说明了。
下面便是单dfs与双dfs的代码
900ms
// UVa 1601 // bfs（原始） #include#include #include #include using namespace std; const int maxc = 20; const int maxn = 256; const int dx[] = {-1, 1, 0, 0, 0}; const int dy[] = {0, 0, -1, 1, 0}; int w, h, n; int s[3], t[3]; int deg[maxn], G[maxn][5]; int get_ID(int a, int b, int c) { return (a<<16)|(b<<8)|c; } bool is_conflict(int a, int b, int a1, int b1) { return (a1 == b1) || (a == b1 && b == a1); } int dist[maxn][maxn][maxn]; int bfs() { queue<int> q; memset(dist, -1, sizeof(dist)); q.push(get_ID(s[0], s[1], s[2])); dist[s[0]][s[1]][s[2]] = 0; while (!q.empty()) { int u = q.front(); q.pop(); int a = (u>>16)&0xff, b = (u>>8)&0xff, c = u&0xff; if (a == t[0] && b == t[1] && c == t[2]) return dist[a][b][c]; for (int i = 0; i < deg[a]; ++i) { int a1 = G[a][i]; for (int j = 0; j < deg[b]; ++j) { int b1 = G[b][j]; if (is_conflict(a, b, a1, b1)) continue; for (int k = 0; k < deg[c]; ++k) { int c1 = G[c][k]; if (is_conflict(a, c, a1, c1)) continue; if (is_conflict(b, c, b1, c1)) continue; if (dist[a1][b1][c1] != -1) continue; dist[a1][b1][c1] = dist[a][b][c] + 1; q.push(get_ID(a1, b1, c1)); } } } } return -1; } int main() { while (scanf("%d%d%d\n", &w, &h, &n) == 3 && n) { int cnt = 0, x[maxn], y[maxn], id[maxc][maxc]; char input[maxc][maxc]; for (int i = 0; i < h; ++i) fgets(input[i], maxc, stdin); for (int i = 0; i < h; ++i) for (int j = 0; j < w; ++j) if (input[i][j] != '#') { x[cnt] = i; y[cnt] = j; id[i][j] = cnt; if (islower(input[i][j])) s[input[i][j]-'a'] = cnt; else if (isupper(input[i][j])) t[input[i][j]-'A'] = cnt; cnt++; } for (int i = 0; i < cnt; ++i) { deg[i] = 0; for (int j = 0; j < 5; ++j) { int x1 = x[i]+dx[j], y1 = y[i]+dy[j]; if (input[x1][y1] != '#') G[i][deg[i]++] = id[x1][y1]; } } if (n <= 2) { deg[cnt] = 1; G[cnt][0] = cnt; s[2] = t[2] = cnt++; } if (n <= 1) { deg[cnt] = 1; G[cnt][0] = cnt; s[1] = t[1] = cnt++; } printf("%d\n", bfs()); } return 0; }
750ms
// UVa 1601 // 双向bfs #include#include #include #include using namespace std; const int maxc = 20; const int maxn = 256; const int dx[] = {1, -1, 0, 0, 0}; const int dy[] = {0, 0, 1, -1, 0}; int w, h, n; int deg[maxn], G[maxn][5]; int s[3], t[3]; int ID(int a, int b, int c) { return (a<<16)|(b<<8)|c; } bool is_conflict(int a, int b, int a1, int b1) { return (a1 == b1) || (a == b1 && b == a1); } int ds[maxn][maxn][maxn]; int dt[maxn][maxn][maxn]; int dfs() { queue<int> qs, qt; qs.push(ID(s[0], s[1], s[2])); qt.push(ID(t[0], t[1], t[2])); memset(ds, -1, sizeof(ds)); memset(dt, -1, sizeof(dt)); ds[s[0]][s[1]][s[2]] = dt[t[0]][t[1]][t[2]] = 0; while (!qs.empty() && !qt.empty()) { { int u = qs.front(); qs.pop(); int a = (u>>16)&0xff, b = (u>>8)&0xff, c = u&0xff; for (int i = 0; i < deg[a]; ++i) { int a1 = G[a][i]; for (int j = 0; j < deg[b]; ++j) { int b1 = G[b][j]; if (is_conflict(a, b, a1, b1)) continue; for (int k = 0; k < deg[c]; ++k) { int c1 = G[c][k]; if (is_conflict(a, c, a1, c1)) continue; if (is_conflict(b, c, b1, c1)) continue; if (ds[a1][b1][c1] == -1) { ds[a1][b1][c1] = ds[a][b][c] + 1; qs.push(ID(a1, b1, c1)); } if (dt[a1][b1][c1] != -1) return ds[a1][b1][c1] + dt[a1][b1][c1]; } } } } { int u = qt.front(); qt.pop(); int a = (u>>16)&0xff, b = (u>>8)&0xff, c = u&0xff; for (int i = 0; i < deg[a]; ++i) { int a1 = G[a][i]; for (int j = 0; j < deg[b]; ++j) { int b1 = G[b][j]; if (is_conflict(a, b, a1, b1)) continue; for (int k = 0; k < deg[c]; ++k) { int c1 = G[c][k]; if (is_conflict(a, c, a1, c1)) continue; if (is_conflict(b, c, b1, c1)) continue; if (dt[a1][b1][c1] == -1) { dt[a1][b1][c1] = dt[a][b][c] + 1; qt.push(ID(a1, b1, c1)); } if (ds[a1][b1][c1] != -1) return dt[a1][b1][c1] + ds[a1][b1][c1]; } } } } } return -1; } int main() { while (scanf("%d%d%d\n", &w, &h, &n) == 3 && n) { char input[maxc][maxc]; int cnt = 0, x[maxn], y[maxn], id[maxc][maxc]; for (int i = 0; i < h; ++i) fgets(input[i], maxc, stdin); for (int i = 0; i < h; ++i) for (int j = 0; j < w; ++j) if (input[i][j] != '#') { x[cnt] = i; y[cnt] = j; id[i][j] = cnt; if (islower(input[i][j])) s[input[i][j]-'a'] = cnt; else if (isupper(input[i][j])) t[input[i][j]-'A'] = cnt; cnt++; } for (int i = 0; i < cnt; ++i) { deg[i] = 0; for (int j = 0; j < 5; ++j) { int x1 = x[i]+dx[j], y1 = y[i]+dy[j]; if (input[x1][y1] != '#') { G[i][deg[i]++] = id[x1][y1]; } } } if (n <= 2) { deg[cnt] = 1; G[cnt][0] = cnt; s[2] = t[2] = cnt++; } if (n <= 1) { deg[cnt] = 1; G[cnt][0] = cnt; s[1] = t[1] = cnt++; } printf("%d\n", dfs()); } return 0; }
单dfs的查询节点大概是 n ^ len
双dfs的查询节点大概是 2 * n ^ (len / 2)
n, len越大优化程度越高