hdu:4771Stealing Harry Potter's Precious（bfs + 全排列）

题目：hdu:4771Stealing Harry Potter‘s Precious

题目大意：给出n* m的矩阵，代表n * m间room，然后每个房间又有脆弱和坚固之分，分别用‘.‘和‘#‘代替。’@‘代表Dudely所在的起点。

题目说Dudely想要偷Harry的宝物，他知道宝物的位置，但是他的魔法只能穿过脆弱的room。愚蠢的他又想偷完harry所有的宝物，并不在乎如何人出去。问最短移动的步数偷完所有的宝物。没法偷完所有的宝物就输出-1.

解题思路：这里要求的是最短路径问题。可以将各个宝物之间的最短距离和起点到每个宝物之间的最短距离都求出来，然后全排列能产生的路径，最后维护总距离的最小值。注意：如果在前面求最短距离的时候，发现有两个宝物或是宝物与起点是不可达的，那么就说明Dudely取不完所有的宝物，可以直接输出-1，后面的不用判断。

代码：

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

const int N = 105;
const int M = 5;
int n, m, Q;
char map[N][N];
int dist[N][N];
int ans[M][M];     //存放最短距离。
int mm;            //最短总距离
int vis[M];        

const int dir[4][2] = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
const int INF = 0x6ffffff;

struct Precious {

	int x, y;
}precious[M], q[N * N];

int Min (const int x , const int y) {return x < y ? x: y;}
int bfs (const Precious begin, const Precious end) {

	memset (dist, 0, sizeof (dist));
	int front, rear;
	front = 0;
	rear = 1;
	q[front].x = begin.x;
	q[front].y = begin.y;
	while (front < rear) {

		if (q[front].x == end.x && q[front].y == end.y)
			return dist[q[front].x][q[front].y];
		for (int i = 0; i < 4; i++) {

			q[rear].x = q[front].x + dir[i][0];
			q[rear].y = q[front].y + dir[i][1];
			if (q[rear].x < 0 || q[rear].x >= n || q[rear].y < 0 || q[rear].y >= m)
				continue;
			if ( map[q[rear].x][q[rear].y] != '#' && !dist[q[rear].x][q[rear].y]) {

				dist[q[rear].x][q[rear].y] = dist[q[front].x][q[front].y] + 1;
				rear++;
			}
		}
		front++;
	}
	return -1;
}

bool solve () {

	memset (ans, -1, sizeof (ans));
	for (int i = 0; i <= Q; i++)
		for (int j = i + 1; j <= Q; j++) {

			ans[i][j] = ans[j][i] = bfs (precious[i], precious[j]);
			if (ans[i][j] == -1)
				return false;
		}

	return true;
}
/*
void dfs (int k, int x, int sum) {

	if (k == Q) {

		mm = Min (mm, sum);
		return;
	}

	for (int i = 1; i <= Q; i++) {

		if ( !vis[i] && ans[x][i]!= -1) {

			if (sum + ans[x][i] >= mm)
				return;
			vis[i] = 1;
			dfs (k + 1, i, sum + ans[x][i]);
			vis[i] = 0;
		}
	}
}
*/
void dfs () {

	int s[M];
	for (int i = 1; i <= Q; i++)
		s[i - 1] = i;
	sort (s, s + Q);
	int sum = 0;
	do {

		sum = ans[0][s[0]];
		for (int i = 0; i < Q - 1; i++)
			sum += ans[s[i]][s[i + 1]];
		mm = Min (mm, sum);
	} while (next_permutation (s, s + Q));
}

int main () {

	char ch;
	while (scanf("%d%d%c", &n, &m, &ch), n || m) {

		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {

				scanf("%c", &map[i][j]);
				if (map[i][j] == '@') {

					precious[0].x = i;
					precious[0].y = j;
				}
			}
			scanf ("%c", &ch);
		}

		scanf("%d", &Q);
		for (int i = 1; i <= Q; i++) {

			scanf ("%d%d", &precious[i].x, &precious[i].y);
			precious[i].x--;
			precious[i].y--;
		}

		if (!solve())
			printf ("-1\n");
		else {

			mm = INF;
			memset(vis, 0, sizeof (vis));
			//dfs (0, 0, 0);
			dfs();
			printf ("%d\n", mm);
		}

	}
	return 0;
}

hdu:4771Stealing Harry Potter's Precious（bfs + 全排列）