版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/sugarbliss/article/details/89388700
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4771
题意：给一个n*m的矩阵，小偷从@出发可以走“.”，墙是“#”不可走，之后又K个珠宝，并且给你这K个珠宝的坐标，问小偷最少要走多少步可以吧珠宝偷完，如果不能偷完输出“-1”。
思路：算上起点"@"一共k+1个点，最终步数的多少一定跟取宝物的顺序有关，我们可以求出k+1个点任意两点的距离，然后dfs搜索一遍即可，也可以用next_permutation全排列函数，其实都一样，下面给出两个版本的代码。
bfs + dfs版本：
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define IOS cin.tie(0); cout.tie(0); ios::sync_with_stdio(0);
const int inf = 0x3f3f3f3f;
const int N = 3e5+7;
int dir[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};
struct node
{
int x, y, step;
}Now, Next, p;
int n, m, k, din[5][505], vis[505][505];
char mp[505][505];
int bfs(int s, int t)
{
memset(vis, 0, sizeof(vis));
p.x = din[s][0]; p.y = din[s][1];
p.step = 0; vis[p.x][p.y] = 1;
queue <node> Q;
Q.push(p);
while(!Q.empty())
{
Now = Q.front(); Q.pop();
if(Now.x == din[t][0] && Now.y == din[t][1]) return Now.step;
for(int i = 0; i < 4; i++)
{
int tx = Now.x + dir[i][0];
int ty = Now.y + dir[i][1];
if(!vis[tx][ty] && tx >= 0 && tx < n && ty >= 0 && ty < m && mp[tx][ty] != '#')
{
Next.x = tx; Next.y = ty;
Next.step = Now.step + 1;
vis[tx][ty] = 1;
Q.push(Next);
}
}
}
return inf;
}
int ans, book[10], dis[10][10];
void dfs(int u, int lev, int sum)
{
if(lev == k)
{
ans = min(ans, sum);
return ;
}
for(int i = 1; i <= k; i++)
{
if(!book[i])
{
book[i] = 1;
dfs(i, lev + 1, sum + dis[u][i]);
book[i] = 0;
}
}
}
int main()
{
while(~scanf("%d%d",&n,&m) && n+m)
{
ans = inf;
memset(din, 0, sizeof(din));
memset(dis, 0x3f, sizeof(dis));
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++)
if(mp[i][j] == '@')
{
din[0][0] = i, din[0][1] = j;
break;
}
scanf("%d",&k);
int a[5] = {0}; int x, y;
for(int i = 1; i <= k; i++)
{
scanf("%d%d",&x, &y);
din[i][0] = --x, din[i][1] = --y;
a[i] = i;
}
for(int i = 0; i < k; i++)
for(int j = i + 1; j <= k; j++)
dis[i][j] = dis[j][i] = bfs(i, j);
memset(book, 0, sizeof(book));
dfs(0, 0, 0);
if(ans == inf) puts("-1");
else printf("%d\n",ans);
}
}
/*
2 3
##@
#.#
1
2 2
4 4
#@##
....
####
....
2
2 1
2 4
*/
bfs + next_permutation版本：
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define IOS cin.tie(0); cout.tie(0); ios::sync_with_stdio(0);
const int inf = 0x3f3f3f3f;
const int N = 3e5+7;
int dir[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};
struct node
{
int x, y, step;
}Now, Next, p;
int n, m, k, din[5][505], vis[505][505];
char mp[505][505];
int bfs(int s, int t)
{
memset(vis, 0, sizeof(vis));
p.x = din[s][0]; p.y = din[s][1];
p.step = 0; vis[p.x][p.y] = 1;
queue <node> Q;
Q.push(p);
while(!Q.empty())
{
Now = Q.front(); Q.pop();
if(Now.x == din[t][0] && Now.y == din[t][1]) return Now.step;
for(int i = 0; i < 4; i++)
{
int tx = Now.x + dir[i][0];
int ty = Now.y + dir[i][1];
if(!vis[tx][ty] && tx >= 0 && tx < n && ty >= 0 && ty < m && mp[tx][ty] != '#')
{
Next.x = tx; Next.y = ty;
Next.step = Now.step + 1;
vis[tx][ty] = 1;
Q.push(Next);
}
}
}
return inf;
}
int main()
{
while(~scanf("%d%d",&n,&m) && n+m)
{
memset(din, 0, sizeof(din));
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++)
if(mp[i][j] == '@')
{
din[0][0] = i, din[0][1] = j;
break;
}
scanf("%d",&k);
int a[5] = {0}; int x, y;
for(int i = 1; i <= k; i++)
{
scanf("%d%d",&x, &y);
din[i][0] = --x, din[i][1] = --y;
a[i] = i;
}
int ans = inf, sum;
do
{
sum = 0;
for(int i = 1; i <= k; i++)
{
int t = bfs(a[i-1], a[i]);
sum += t;
}
ans = min(ans, sum);
}while(next_permutation(a+1, a+1+k));
if(ans == inf) puts("-1");
else printf("%d\n",ans);
}
}
/*
2 3
##@
#.#
1
2 2
4 4
#@##
....
####
....
2
2 1
2 4
*/