hdu 4771 Stealing Harry Potter's Precious(bfs)
题目链接:hdu 4771 Stealing Harry Potter's Precious
题目大意:在一个N*M的银行里,贼的位置在’@‘,如今给出n个宝物的位置。如今贼要将全部的宝物拿到手。问最短的路径,不须要考虑离开。
解题思路:由于宝物最多才4个,加上贼的位置,枚举两两位置,用bfs求两点距离,假设存在两点间不能到达,那么肯定是不能取全然部的宝物。
然后枚举取宝物的顺序。维护ans最小。
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 100;
const int maxv = 10;
const int INF = 0x3f3f3f3f;
const int dir[4][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };
struct point {
int x, y;
point (int x = 0, int y = 0) {
this->x = x;
this->y = y;
}
} p[maxv];
char g[maxn+5][maxn+5];
int n, N, M, v[maxn+5][maxn+5], d[maxv][maxv];
void init () {
memset(g, 0, sizeof(g));
for (int i = 1; i <= N; i++) {
scanf("%s", g[i] + 1);
for (int j = 1; j <= M; j++)
if (g[i][j] == '@') {
p[0].x = i;
p[0].y = j;
}
}
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d%d", &p[i].x, &p[i].y);
}
int bfs (point s, point e) {
queue<point> que;
memset(v, -1, sizeof(v));
que.push(s);
v[s.x][s.y] = 0;
while (!que.empty()) {
point u = que.front();
que.pop();
if (u.x == e.x && u.y == e.y)
return v[u.x][u.y];
for (int i = 0; i < 4; i++) {
int xi = u.x + dir[i][0];
int yi = u.y + dir[i][1];
if (xi > N || xi <= 0)
continue;
if (yi > M || yi <= 0)
continue;
if (v[xi][yi] != -1 || g[xi][yi] == '#')
continue;
que.push(point(xi, yi));
v[xi][yi] = v[u.x][u.y] + 1;
}
}
return -1;
}
bool judge () {
memset(d, 0, sizeof(d));
for (int i = 0; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
int dis = bfs(p[i], p[j]);
if (dis == -1)
return false;
d[i][j] = d[j][i] = dis;
}
}
return true;
}
int solve () {
int pos[maxv];
for (int i = 0; i <= n; i++)
pos[i] = i;
int ans = INF;
do {
int sum = 0;
for (int i = 0; i < n; i++)
sum += d[pos[i]][pos[i+1]];
ans = min (ans, sum);
} while(next_permutation(pos + 1, pos + n + 1));
return ans;
}
int main () {
while (scanf("%d%d", &N, &M) == 2 && N + M) {
init();
if (judge()) {
printf("%d\n", solve());
} else
printf("-1\n");
}
return 0;
}