题目链接
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3865
思路
一个迷宫题 但是每次的操作数和普通的迷宫题不一样
0.按下按钮 即向当前方向盘的方向前进一格 1.往左移动方向盘 2.往右移动方向盘 3.不动
然后记录时间
易知，可以从四个方向到达终点，最后从这四个方向找最小值就是答案
当然要判断 不能通达的结果
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a) memset(a, 0, sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; typedef pair<string, int> psi; typedef pair<string, string> pss; const double PI = acos(-1); const double E = exp(1); const double eps = 1e-30; const int INF = 0x3f3f3f3f; const int maxn = 5e4 + 5; const int MOD = 1e9 + 7; char dis[4][4] = { 'l', 'r', 'u', 'd', 'd', 'l', 'r', 'u', 'u', 'd', 'l', 'r', 'r', 'u', 'd', 'l', }; map <char, int> M; int Move[4][2] { -1, 0, // up 1, 0, //down 0,-1, // left 0, 1, //right }; char G[15][15]; int n, m, p; int dp[12][12][4]; int ans[12][12][4]; struct Node { int x, y, d; }tmp; queue <Node> q; int sx, sy, ex, ey; bool ok(int x, int y, int d) { if (x < 0 || x >= n || y < 0 || y >= m) return false; if (G[x][y] == '*' || dp[x][y][d] != -1) return false; return true; } void bfs() { while (!q.empty()) { int x = q.front().x; int y = q.front().y; int d = q.front().d; q.pop(); if (x == ex && y == ey && ans[x][y][d] == -1) ans[x][y][d] = dp[x][y][d]; int nx = x + Move[M[dis[dp[x][y][d] / p % 4][d]]][0]; int ny = y + Move[M[dis[dp[x][y][d] / p % 4][d]]][1]; // press botton if (ok(nx, ny, d)) { dp[nx][ny][d] = dp[x][y][d] + 1; tmp.x = nx; tmp.y = ny; tmp.d = d; q.push(tmp); } // move left int nd = d - 1; if (nd < 0) nd = 3; if (ok(x, y, nd)) { dp[x][y][nd] = dp[x][y][d] + 1; tmp.x = x; tmp.y = y; tmp.d = nd; q.push(tmp); } // move right nd = d + 1; if (nd > 3) nd = 0; if (ok(x, y, nd)) { dp[x][y][nd] = dp[x][y][d] + 1; tmp.x = x; tmp.y = y; tmp.d = nd; q.push(tmp); } // stay if (dp[x][y][d] <= 200) { dp[x][y][d]++; tmp.x = x; tmp.y = y; tmp.d = d; q.push(tmp); } } } void init() { M['u'] = 0; M['d'] = 1; M['l'] = 2; M['r'] = 3; } int main() { init(); int T; cin >> T; while (T--) { memset(dp, -1, sizeof(dp)); memset(ans, -1, sizeof(ans)); scanf("%d%d%d", &n, &m, &p); int x = -1, y = -1; for (int i = 0; i < n; i++) { scanf("%s", &G[i]); for (int j = 0; j < m; j++) { if (G[i][j] == '@') { sx = i; sy = j; } else if (G[i][j] == '$') { ex = i; ey = j; } } } dp[sx][sy][0] = 0; tmp.x = sx; tmp.y = sy; tmp.d = 0; q.push(tmp); bfs(); int Ans = INF; for (int i = 0; i < 4; i++) { if (ans[ex][ey][i] != -1) Ans = min(Ans, ans[ex][ey][i]); } if (Ans != INF) printf("%d\n", Ans); else printf("YouBadbad\n"); } }转载于:https://www.cnblogs.com/Dup4/p/9433142.html
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