hdu 2312 Cliff Climbing (pfs)
一条很暴力，有点恶心的搜索。题意其实很简单，主要是pfs的时候拓展结点会有种麻烦的感觉。注意的是，这里的n和m跟平常见到的有所不同，交换过来了。我的代码就是在因为这个长宽的问题wa了一次。
代码如下：
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cstring> 5 #include <queue> 6 #include <cctype> 7 8 using namespace std; 9 10 typedef pair<int, int> PII; 11 typedef pair<PII, PII> PPP; 12 typedef pair<int, PPP> PPPI; 13 14 #define FI first 15 #define SE second 16 #define PRIQ priority_queue 17 18 const int R = 66; 19 const int C = 33; 20 char mat[R][C]; 21 bool vis[R][C][R][C]; 22 int n, m; 23 24 inline int dis(int a, int b, int c, int d) { return abs(a - c) + abs(b - d);} 25 inline bool inmat(int a, int b) { return 0 <= a && a < n && 0 <= b && b < m;} 26 27 int work() { 28 PRIQ<PPPI> Q; 29 int nx, ny, cx, cy; 30 memset(vis, 0, sizeof(vis)); 31 while (!Q.empty()) Q.pop(); 32 for (int i = 0; i < n; i++) { 33 for (int j = 0; j < m; j++) { 34 if (mat[i][j] != 'S') continue; 35 for (int dx = -2; dx <= 2; dx++) { 36 nx = i + dx; 37 for (int dy = 1; dy <= 3; dy++) { 38 ny = j + dy; 39 if (inmat(nx, ny) && dis(i, j, nx, ny) <= 3 && mat[nx][ny] != 'X' && !vis[i][j][nx][ny]) { 40 if (isdigit(mat[nx][ny])) Q.push(PPPI((int) '0' - mat[nx][ny], PPP(PII(i, j), PII(nx, ny)))); 41 else Q.push(PPPI(0, PPP(PII(i, j), PII(nx, ny)))); 42 vis[i][j][nx][ny] = true; 43 // cout << i << ' ' << j << ' ' << nx << ' ' << ny << endl; 44 } 45 ny = j - dy; 46 if (inmat(nx, ny) && dis(nx, ny, i, j) <= 3 && mat[nx][ny] != 'X' && !vis[nx][ny][i][j]) { 47 if (isdigit(mat[nx][ny])) Q.push(PPPI((int) '0' - mat[nx][ny], PPP(PII(nx, ny), PII(i, j)))); 48 else Q.push(PPPI(0, PPP(PII(nx, ny), PII(i, j)))); 49 vis[nx][ny][i][j] = true; 50 // cout << nx << ' ' << ny << ' ' << i << ' ' << j << endl; 51 } 52 } 53 } 54 } 55 } 56 // cout << "find S" << endl; 57 while (!Q.empty()) { 58 PPPI cur = Q.top(); 59 Q.pop(); 60 int v = cur.FI, a = cur.SE.FI.FI, b = cur.SE.FI.SE, c = cur.SE.SE.FI, d = cur.SE.SE.SE; 61 // cout << v << ' ' << a << ' ' << b << ' ' << c << ' ' << d << endl; 62 if (mat[a][b] == 'T' || mat[c][d] == 'T') return -v; 63 for (int dx = -2; dx <= 2; dx++) { 64 nx = a + dx; 65 cx = c + dx; 66 for (int dy = 1; dy <= 3; dy++) { 67 ny = b + dy; 68 if (inmat(nx, ny) && dis(a, b, nx, ny) <= 3 && mat[nx][ny] != 'X' && !vis[a][b][nx][ny]) { 69 if (isdigit(mat[nx][ny])) Q.push(PPPI((int) '0' - mat[nx][ny] + v, PPP(PII(a, b), PII(nx, ny)))); 70 else Q.push(PPPI(v, PPP(PII(a, b), PII(nx, ny)))); 71 vis[a][b][nx][ny] = true; 72 // cout << nx << ' ' << ny << endl; 73 } 74 cy = d - dy; 75 if (inmat(cx, cy) && dis(cx, cy, c, d) <= 3 && mat[cx][cy] != 'X' && !vis[cx][cy][c][d]) { 76 if (isdigit(mat[cx][cy])) Q.push(PPPI((int) '0' - mat[cx][cy] + v, PPP(PII(cx, cy), PII(c, d)))); 77 else Q.push(PPPI(v, PPP(PII(cx, cy), PII(c, d)))); 78 vis[cx][cy][c][d] = true; 79 // cout << cx << ' ' << cy << endl; 80 } 81 } 82 } 83 } 84 return -1; 85 } 86 87 int main() { 88 // freopen("in", "r", stdin); 89 // freopen("out", "w", stdout); 90 while (cin >> m >> n && (n || m)) { 91 char buf[3]; 92 for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) cin >> buf, mat[i][j] = buf[0]; 93 cout << work() << endl; 94 } 95 return 0; 96 }
——written by Lyon
