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题意:一个人从(0, 0)逃往(n, m),地图上有朝某个方向开炮的炮台,问最少逃脱步数
分析:主要在状态是否OK,当t时刻走到(x,y),炮台是否刚好打中,因为只能是整数,所以用整除判断。题意不清楚,有些坑点。
#include <bits/stdc++.h> using namespace std; const int N = 1e2 + 5; struct Point { int dir, t, v; //N 1 E 2 S 3 W 4 }p[N][N]; struct Node { int x, y, step; Node() {} Node(int x, int y, int step) : x (x), y (y), step (step) {} }; int dx[5] = {-1, 1, 0, 0, 0}; int dy[5] = {0, 0, -1, 1, 0}; bool vis[N][N][N*10]; int n, m, k, d; bool check(int x, int y, int tim) { if (x < 0 || x > n || y < 0 || y > m || vis[x][y][tim] || p[x][y].dir != 0) return false; else return true; } bool check2(int x, int y, int tim) { for (int i=x-1; i>=0; --i) { //up if (p[i][y].v == 0) continue; if (p[i][y].dir != 3) break; int dis = x - i; if (dis % p[i][y].v != 0) break; int t = tim - dis / p[i][y].v; if (t % p[i][y].t == 0) return false; else break; } for (int i=x+1; i<=n; ++i) { //down if (p[i][y].v == 0) continue; if (p[i][y].dir != 1) break; int dis = i - x; if (dis % p[i][y].v != 0) break; int t = tim - dis / p[i][y].v; if (t % p[i][y].t == 0) return false; else break; } for (int i=y-1; i>=0; --i) { //left if (p[x][i].v == 0) continue; if (p[x][i].dir != 2) break; int dis = y - i; if (dis % p[x][i].v != 0) break; int t = tim - dis / p[x][i].v; if (t % p[x][i].t == 0) return false; else break; } for (int i=y+1; i<=m; ++i) { //right if (p[x][i].v == 0) continue; if (p[x][i].dir != 4) break; int dis = i - y; if (dis % p[x][i].v != 0) break; int t = tim - dis / p[x][i].v; if (t % p[x][i].t == 0) return false; else break; } return true; } int BFS(void) { Node s; s.x = s.y = s.step = 0; queue<Node> que; que.push (s); vis[s.x][s.y][0] = true; while (!que.empty ()) { Node u = que.front (); que.pop (); if (u.step > d) continue; if (u.x == n && u.y == m && u.step <= d) { return u.step; } for (int i=0; i<5; ++i) { int tx = u.x + dx[i], ty = u.y + dy[i], tsp = u.step + 1; if (!check (tx, ty, tsp)) continue; if (!check2 (tx, ty, tsp)) continue; vis[tx][ty][tsp] = true; que.push (Node (tx, ty, tsp)); } } return -1; } int main(void) { map<char, int> mp; mp[‘N‘] = 1; mp[‘E‘] = 2; mp[‘S‘] = 3; mp[‘W‘] = 4; while (scanf ("%d%d%d%d", &n, &m, &k, &d) == 4) { memset (vis, false, sizeof (vis)); memset (p, 0, sizeof (p)); char str[3]; int t, v, x, y; getchar (); for (int i=0; i<k; ++i) { scanf ("%s%d%d%d%d", &str, &t, &v, &x, &y); p[x][y].dir = mp[str[0]]; p[x][y].t = t; p[x][y].v = v; } int ans = BFS (); if (ans == -1) puts ("Bad luck!"); else printf ("%d\n", ans); } return 0; }
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原文地址:http://www.cnblogs.com/Running-Time/p/4989833.html