UVA816 Abbott的复仇 Abbott‘s Revenge

主要是有三个点，状态是三元的，包括进入某点时的朝向，然后就是对方向的操作，这里包括怎么处理输入的数据，最后就是要打印任意一条路径，所以要在搜索的中间记录路径，然后写一个打印函数，这个代码明天再修改一下

#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pa;
struct state {
int x, y, dir;
state(int a, int b, int c): x(a), y(b), dir(c) {}
};
int xs, ys, xe, ye;
char dir;
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
int vis[10][10][5];
char str[5] = "NESW";
map<pa, vector<string> > mp;
map<char, int> mp2;
int pre[10][10][5];
int getdir(int a, int b) {
if (b == 'L') {
if (a) a--;
else a = 3;
} else {
if (a == 3) a = 0;
else a++;
}
return a;
}
int bfs() {
queue<state> q;
int r = xs + dx[mp2[dir]];
int c = ys + dy[mp2[dir]];
memset(vis, 0, sizeof(vis));
if (r >= 1 && r <= 9 && c >= 1 && c <= 9) {
q.push(state(r, c, mp2[dir]));
vis[r][c][mp2[dir]] = 1;
pre[r][c][mp2[dir]] = xs * 10 + ys;
}
while (!q.empty()) {
state now = q.front();
q.pop();
if (now.x == xe && now.y == ye) { dir = str[now.dir]; return vis[now.x][now.y][now.dir]; }
pa p = mk(now.x, now.y);
for (int i = 0; i < sz(mp[p]); i++) {
if (mp[p][i][0] == str[now.dir]) {
string t = mp[p][i];
for (int j = 1; j < sz(t); j++) {
int dir_t;
if (t[j] == 'F') dir_t = now.dir;
else dir_t = getdir(now.dir, t[j]);
int x = now.x + dx[dir_t];
int y = now.y + dy[dir_t];
if (x < 1 || x > 9 || y < 1 || y > 9) continue;
if (vis[x][y][dir_t]) continue;
q.push(state(x, y, dir_t));
vis[x][y][dir_t] = vis[now.x][now.y][now.dir] + 1;
pre[x][y][dir_t] = now.x * 100 + now.y * 10 + now.dir;
}
}
}
}
return -1;
}
void print() {
int r = xe, c = ye;
vector<pa> ans;
int z = mp2[dir];
while (true) {
ans.pb(mk(r, c));
if (vis[r][c][z] == 1) break;
int num = pre[r][c][z];
r = num / 100;
c = num % 100 / 10;
z = num % 10;
}
ans.pb(mk(xs, ys));
reverse(all(ans));
for (int i = 0; i < sz(ans); i++) {
if (i % 10 == 0) cout << " ";
cout << " (" << ans[i].fi << "," << ans[i].se << ")";
if (i % 10 == 9 || i == sz(ans) - 1) cout << endl;
}
}
int main() {
mp2['N'] = 0; mp2['E'] = 1; mp2['S'] = 2; mp2['W'] = 3;
string maze;
while (cin >> maze && maze != "END") {
cout << maze << endl;
cin >> xs >> ys >> dir >> xe >> ye;
int x, y;
while (cin >> x && x) {
cin >> y;
string ss;
while (cin >> ss && ss != "*") {
mp[mk(x, y)].pb(ss);
}
}
int rec = bfs();
if (rec == -1) cout << "
No Solution Possiblen";
else print();
mp.clear();
}
return 0;
}

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