旅行商問題，狀壓dp
用一個二進制數表示是否走過當前的幾個位置
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <string>
#include <math.h>
#include <stdlib.h>
using namespace std;
int n,m;
char g[110][110];
int a[110][110];
int sx,sy;
int k;
int dp[110][110][32];
int px[10],py[10];
struct Node
{
int x,y;
int s;
Node(int _x = 0,int _y = 0,int _s = 0)
{
x = _x;
y = _y;
s = _s;
}
};
int move_[][2] = {{0,1},{0,-1},{1,0},{-1,0}};
int bfs()
{
queue<Node>q;
int s = 0;
for(int i = 0;i < k;i++)
if(sx == px[i] && sy == py[i])
s = s|(1<<i);
q.push(Node(sx,sy,s));
memset(dp,-1,sizeof(dp));
dp[sx][sy][s] = 0;
while(!q.empty())
{
Node tmp = q.front();
q.pop();
if(tmp.s == ((1<<k) - 1))
{
return dp[tmp.x][tmp.y][tmp.s];
}
for(int i = 0;i < 4;i++)
{
int nx = tmp.x + move_[i][0];
int ny = tmp.y + move_[i][1];
int s = tmp.s;
if(nx < 0 || nx >= n || ny < 0 || ny >= m)continue;
if(a[nx][ny] == -2)continue;
for(int j = 0;j < k;j++)
if(nx == px[j] && ny == py[j])
{
s |= (1<<j);
}
if(dp[nx][ny][s] != -1)continue;
dp[nx][ny][s] = dp[tmp.x][tmp.y][tmp.s] + 1;
q.push(Node(nx,ny,s));
}
}
return -1;
}
int main()
{
while(scanf("%d%d",&n,&m) == 2)
{
if(n == 0 && m == 0)break;
for(int i = 0;i < n;i++)
scanf("%s",g[i]);
memset(a,-1,sizeof(a));
for(int i = 0;i < n;i++)
for(int j = 0;j < m;j++)
{
if(g[i][j] == '@')
{
sx = i;
sy = j;
}
if(g[i][j] == '#')
a[i][j] = -2;
}
scanf("%d",&k);
int x,y;
for(int i = 0;i < k;i++)
{
scanf("%d%d",&x,&y);
x --;
y--;
px[i] = x;
py[i] = y;
}
printf("%d\n",bfs());
}
return 0;
}