HDU 1044 bfs+dfs

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#include <cstdio>#include <cstring>#include <cctype>#include <queue>using namespace std;#define check(x,y) x>=0&&x<n&&y>=0&&y<m&&!vis[x][y]&&mp[x][y]!='*'const int maxn = 50 + 10;const int dir[4][2] = { -1, 0, 1, 0, 0, -1, 0, 1};int n, m, T, w, t, kase, val[maxn], dis[maxn][maxn], step[maxn][maxn], ans, sum;bool vis[maxn][maxn], res[maxn];char mp[maxn][maxn];void bfs(int x, int y, int s){queue<int>Q;Q.push(x * m + y);vis[x][y] = 1, step[x][y] = 0;while (!Q.empty()){int cur = Q.front(); Q.pop();x = cur / m; y = cur % m;for (int i = 0; i < 4; i++){int Tx = x + dir[i][0], Ty = y + dir[i][1];if (check(Tx, Ty)){vis[Tx][Ty] = 1;step[Tx][Ty] = step[x][y] + 1;if (mp[Tx][Ty] == '@') dis[s][0] = step[Tx][Ty];if (isalpha(mp[Tx][Ty])) dis[s][mp[Tx][Ty] - '@'] = step[Tx][Ty];if (mp[Tx][Ty] == '<') dis[s][w + 1] = step[Tx][Ty];Q.push(Tx * m + Ty);}}}}void dfs(int p, int s, int Time){if (Time > t || ans == sum) return;if (p > w && s > ans) ans = s;for (int i = 0; i <= w + 1; i++)if (dis[p][i] && !res[i]){res[i] = 1;dfs(i, s + val[i], Time + dis[p][i]);res[i] = 0;}}int main(int argc, char const *argv[]){scanf("%d", &T);while (T--){scanf("%d%d%d%d", &m, &n, &t, &w);sum = 0, ans = -1; val[0] = val[w + 1] = 0;memset(vis, 0, sizeof(vis));memset(res, 0, sizeof(res));memset(dis, 0, sizeof(dis));for (int i = 1; i <= w; i++)scanf("%d", &val[i]), sum += val[i];for (int i = 0; i < n; i++)scanf("%s", &mp[i]);for (int i = 0; i < n; i++)for (int j = 0; j < m; j++){memset(step, 0, sizeof(step));memset(vis, 0, sizeof(vis));if (mp[i][j] == '@') bfs(i, j, 0);if (isalpha(mp[i][j])) bfs(i, j, mp[i][j] - '@');if (mp[i][j] == '<') bfs(i, j, w + 1);}res[0] = 1;dfs(0, 0, 0);printf("Case %d:\n", ++kase);if (ans >=  0) printf("The best score is %d.\n", ans);else printf("Impossible\n");if (T) printf("\n");}return 0;}

迷宫中，起点到终点，有几个宝物，在给定的时间内，到达终点且能取得多少价值的宝物。

需要解决4个问题：

1、求入口到第一个取宝物的地方的最短距离
2、求第i个取宝物的地方到第i+1个取宝物的地方的最短距离
3、求第n个取宝物的地方到出口的最短距离
4、保证以上3点能在时间L内实现的情况下，取得的宝石价值最大

用bfs解决1-3，dfs求答案。

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