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HDU 4771 Stealing Harry Potter's Precious(BFS + DFS)

2014-07-06 21:00 519 查看


HDU 4771 Stealing Harry Potter's Precious

题目链接

题意:给定人的起始位置和k个宝物,求人拿完全部宝物最小的步数

思路:先bfs打出两两之间路径,然后dfs暴力求答案,因为宝物才4个,所以暴力是没问题的

代码:

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;

const int INF = 0x3f3f3f3f;
const int d[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
const int N = 105;
int n, m, k, g[10][10], vis[10];
char str

;

struct Point {
int x, y, num;
Point(int x = 0, int y = 0) {
this->x = x;
this->y = y;
}
} s, p[10];

int bfs(Point a, Point b) {
queue<Point> Q;
a.num = 0;
Q.push(a);
int vis

;
memset(vis, 0, sizeof(vis));
vis[a.x][a.y] = 1;
while (!Q.empty()) {
Point now = Q.front();
if (now.x == b.x && now.y == b.y) return now.num;
Q.pop();
for (int i = 0; i < 4; i++) {
int xx = now.x + d[i][0];
int yy = now.y + d[i][1];
if (xx <= 0 || xx > n || yy <= 0 || yy > m || str[xx][yy] == '#'|| vis[xx][yy]) continue;
Point next = Point(xx, yy);
next.num = now.num + 1;
vis[next.x][next.y] = 1;
Q.push(next);
}
}
return -1;
}

bool build() {
memset(g, -1, sizeof(g));
for (int i = 0; i < k; i++) {
g[0][i + 1] = bfs(s, p[i]);
for (int j = i + 1; j < k; j++)
g[i + 1][j + 1] = g[j + 1][i + 1] = bfs(p[i], p[j]);
}
return true;
}

int dfs(int now, int num) {
if (num == k) return 0;
int ans = INF;
for (int i = 1; i <= k; i++) {
if (g[now][i] == -1 || vis[i]) continue;
vis[i] = 1;
ans = min(ans, dfs(i, num + 1) + g[now][i]);
vis[i] = 0;
}
return ans;
}

int main() {
while (~scanf("%d%d", &n, &m) && n || m) {
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= n; i++) {
scanf("%s", str[i] + 1);
for (int j = 1; j <= m; j++) {
if (str[i][j] == '@')
s = Point(i, j);
}
}
scanf("%d", &k);
for (int i = 0; i < k; i++)
scanf("%d%d", &p[i].x, &p[i].y);
if (!build()) printf("-1\n");
else printf("%d\n", dfs(0, 0));
}
return 0;
}
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