UVA 816 - Abbott's Revenge(BFS)

UVA 816 - Abbott's Revenge

option=com_onlinejudge&Itemid=8&page=show_problem&category=599&problem=757&mosmsg=Submission+received+with+ID+14332151" target="_blank" style="">题目链接

题意:一个迷宫,每一个点限制了从哪一方向来的。仅仅能往左右前走,然后问起点到终点的最短路径

思路:BFS。每一个点拆成4个方向的点。相应能走的方向建图跑一下bfs就可以

代码:

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std; const int N = 10005;
const int D[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};
char name[25]; int n, m;
vector<int> g[15][15][4]; struct State {
int x, y, dir;
int pre;
} Q[N], s, e; char str[25];
int x, y, vis[15][15][4]; int hash(char c) {
if (c == 'F') return 0;
if (c == 'R') return 1;
if (c == 'L') return -1;
if (c == 'N') return 0;
if (c == 'E') return 1;
if (c == 'S') return 2;
return 3;
} #define MP(a,b) make_pair(a,b)
typedef pair<int, int> pii;
vector<pii> ans; void print(int u) {
if (u == -1) return;
print(Q[u].pre);
ans.push_back(MP(Q[u].x, Q[u].y));
} void bfs() {
ans.clear();
memset(vis, 0, sizeof(vis));
int head = 0, rear = 0; s.pre = -1;
Q[rear++] = s;
vis[s.x][s.y][s.dir] = 1;
while (head < rear) {
State u = Q[head++];
if (u.x == e.x && u.y == e.y) {
print(head - 1);
int tot = ans.size();
for (int i = 0; i < tot; i++) {
if (i % 10 == 0) printf("\n ");
printf(" (%d,%d)", ans[i].first, ans[i].second);
}
printf("\n");
return;
}
for (int i = 0; i < g[u.x][u.y][u.dir].size(); i ++) {
int di = (g[u.x][u.y][u.dir][i] + u.dir + 4) % 4;
State v = u;
v.x += D[di][0]; v.y += D[di][1];
if (v.x < 0 || v.y < 0) continue;
v.dir = di;
if (vis[v.x][v.y][v.dir]) continue;
vis[v.x][v.y][v.dir] = 1;
v.pre = head - 1;
Q[rear++] = v;
}
}
printf("\n No Solution Possible\n");
} int main() {
while (~scanf("%s", name) && strcmp(name, "END")) {
memset(g, 0, sizeof(g));
printf("%s", name);
scanf("%d%d%s", &s.x, &s.y, str);
s.dir = hash(str[0]);
scanf("%d%d", &e.x, &e.y);
g[s.x][s.y][hash(str[0])].push_back(0);
int x, y;
while (scanf("%d", &x) && x) {
scanf("%d", &y);
while (scanf("%s", str) && str[0] != '*') {
int len = strlen(str);
for (int i = 1; i < len; i++)
g[x][y][hash(str[0])].push_back(hash(str[i]));
}
}
bfs();
}
return 0;
}
上一篇:Codeforces#371 Div2


下一篇:新做的一个基于OPENGL的gui库