Hdu 4856 Tunnels（状压dp）
题目链接
Tunnels
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1247 Accepted Submission(s): 378
Problem Description
Bob is travelling in Xi’an. He finds many secret tunnels beneath the city. In his eyes, the city is a grid. He can’t enter a grid with a barrier. In one minute, he can move into an adjacent grid with no barrier. Bob is full of curiosity and he wants to visit all of the secret tunnels beneath the city. To travel in a tunnel, he has to walk to the entrance of the tunnel and go out from the exit after a fabulous visit. He can choose where he starts and he will travel each of the tunnels once and only once. Now he wants to know, how long it will take him to visit all the tunnels (excluding the time when he is in the tunnels).
Input
The input contains mutiple testcases. Please process till EOF.
For each testcase, the first line contains two integers N (1 ≤ N ≤ 15), the side length of the square map and M (1 ≤ M ≤ 15), the number of tunnels.
The map of the city is given in the next N lines. Each line contains exactly N characters. Barrier is represented by “#” and empty grid is represented by “.”.
Then M lines follow. Each line consists of four integers x 1, y 1, x 2, y 2, indicating there is a tunnel with entrence in (x 1, y 1) and exit in (x 2, y 2). It’s guaranteed that (x 1, y 1) and (x 2, y 2) in the map are both empty grid.
For each testcase, the first line contains two integers N (1 ≤ N ≤ 15), the side length of the square map and M (1 ≤ M ≤ 15), the number of tunnels.
The map of the city is given in the next N lines. Each line contains exactly N characters. Barrier is represented by “#” and empty grid is represented by “.”.
Then M lines follow. Each line consists of four integers x 1, y 1, x 2, y 2, indicating there is a tunnel with entrence in (x 1, y 1) and exit in (x 2, y 2). It’s guaranteed that (x 1, y 1) and (x 2, y 2) in the map are both empty grid.
Output
For each case, output a integer indicating the minimal time Bob will use in total to walk between tunnels.
If it is impossible for Bob to visit all the tunnels, output -1.
If it is impossible for Bob to visit all the tunnels, output -1.
Sample Input
5 4 ....# ...#. ..... ..... ..... 2 3 1 4 1 2 3 5 2 3 3 1 5 4 2 1
Sample Output
7
Source
2014西安全国邀请赛
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题意：n*n的网格图，‘.’ 可以走，‘#’不可以走。有m条隧道，每条隧道有一个起点、终点。从一个点走到相邻的点要1分钟。求访问完所有隧道的最短时间，一条隧道只能访问一次。
题解：由于m很小，我们可以状压。直接在原图上跑状压复杂度太高。所以，可以先求出连在隧道上的点之间的距离，在这些点上跑状压就行了（注意一些小的优化）。代码如下：
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<math.h>
#include<string>
#include<set>
#include<map>
#define nn 610000
#define inff 0x3fffffff
using namespace std;
typedef __int64 LL;
typedef unsigned __int64 ULL;
int n,m;
int dir[4][2]={1,0,-1,0,0,1,0,-1};
char tu[25][25];
struct node
{
int x,y;
node(){};
node(int xx,int yy)
{
x=xx,y=yy;
}
}u[40];
struct bian
{
int en,len,next;
int z;
}E[40*40*2];
int p[40],num;
void init()
{
memset(p,-1,sizeof(p));
num=0;
}
void add(int st,int en,int len,int ty)
{
E[num].en=en;
E[num].len=len;
E[num].next=p[st];
E[num].z=ty;
p[st]=num++;
}
int dis[20][20];
queue<node>que;
void bfs(int x,int y)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
dis[i][j]=inff;
}
}
dis[x][y]=0;
que.push(node(x,y));
node sta;
int dx,dy;
while(que.size())
{
sta=que.front();
que.pop();
for(i=0;i<4;i++)
{
dx=sta.x+dir[i][0],dy=sta.y+dir[i][1];
if(dx>=0&&dx<n&&dy>=0&&dy<n&&tu[dx][dy]!='#')
{
if(dis[dx][dy]==inff)
{
dis[dx][dy]=dis[sta.x][sta.y]+1;
que.push(node(dx,dy));
}
}
}
}
}
struct so
{
int id,zt;
so(){}
so(int x,int y)
{
id=x,zt=y;
}
};
int dp[40][(1<<15)+10];
bool inque[40][(1<<15)+10];
queue<so>q;
int ans;
void solve()
{
ans=inff;
int i,j;
for(i=0;i<=2*m;i++)
{
for(j=0;j<(1<<m);j++)
{
dp[i][j]=inff;
inque[i][j]=false;
}
}
dp[0][0]=0;
q.push(so(0,0));
so sta;
int w,zt;
while(q.size())
{
sta=q.front();
q.pop();
//cout<<sta.id<<" "<<sta.zt<<" "<<dp[sta.id][sta.zt]<<endl;
inque[sta.id][sta.zt]=false;
if(sta.zt==(1<<m)-1)
{
ans=min(ans,dp[sta.id][sta.zt]);
continue;
}
for(i=p[sta.id];i+1;i=E[i].next)
{
w=E[i].en;
zt=sta.zt;
if(E[i].z>=0)
{
if(((1<<E[i].z)&zt)==0)
zt+=(1<<E[i].z);
else
continue;
}
if(dp[w][zt]>dp[sta.id][sta.zt]+E[i].len)
{
dp[w][zt]=dp[sta.id][sta.zt]+E[i].len;
if(!inque[w][zt])
{
inque[w][zt]=true;
q.push(so(w,zt));
}
}
}
}
}
int main()
{
int i,j;
int xx,x,y,yy;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<n;i++)
{
scanf("%s",tu[i]);
}
int cnt=0;
for(i=0;i<m;i++)
{
scanf("%d%d%d%d",&x,&y,&xx,&yy);
x--,y--,xx--,yy--;
u[++cnt].x=x;
u[cnt].y=y;
u[++cnt].x=xx;
u[cnt].y=yy;
}
init();
for(i=1;i<=cnt;i++)
{
bfs(u[i].x,u[i].y);
if(i%2)
{
add(0,i,0,-1);
add(i,i+1,0,i/2);
}
else
for(j=1;j<=cnt;j+=2)
{
if(i==j)
continue;
add(i,j,dis[u[j].x][u[j].y],-1);
}
}
solve();
if(ans==inff)
ans=-1;
printf("%d\n",ans);
}
return 0;
}
仔细分析，我们可以发现，我们已经求出了点与点之间的最小距离，换句话说一个点只会访问一次，到了隧道的起点一定要走这条隧道。所以，我们可以把隧道看成一个点，把一条隧道的终点到另一条隧道的起点的距离，看成这条隧道到另一条隧道的距离。这样问题就转换成了经典的TSP问题。代码如下：
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<math.h>
#include<string>
#include<set>
#include<map>
#define nn 610000
#define inff 0x3fffffff
using namespace std;
int n,m;
char tu[20][20];
struct node
{
int x,y;
node(){}
node(int xx,int yy)
{
x=xx,y=yy;
}
}u[20],v[20];
struct bian
{
int en,next,len;
}E[20*20*2];
int p[20],num;
void init()
{
memset(p,-1,sizeof(p));
num=0;
}
void add(int st,int en,int len)
{
E[num].en=en;
E[num].len=len;
E[num].next=p[st];
p[st]=num++;
}
queue<node>que;
int dis[20][20];
int dir[4][2]={1,0,-1,0,0,1,0,-1};
void bfs(int x,int y)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
dis[i][j]=inff;
}
}
dis[x][y]=0;
que.push(node(x,y));
node sta;
int dx,dy;
while(que.size())
{
sta=que.front();
que.pop();
for(i=0;i<4;i++)
{
dx=sta.x+dir[i][0],dy=sta.y+dir[i][1];
if(dx>=0&&dx<n&&dy>=0&&dy<n&&tu[dx][dy]=='.')
{
if(dis[dx][dy]==inff)
{
dis[dx][dy]=dis[sta.x][sta.y]+1;
que.push(node(dx,dy));
}
}
}
}
}
int dp[20][(1<<15)+10];
void solve()
{
int i,j,k;
for(i=0;i<=m;i++)
{
for(j=0;j<(1<<m);j++)
{
dp[i][j]=inff;
}
}
dp[0][0]=0;
int w,zt;
for(j=0;j<(1<<m);j++)
{
for(i=0;i<=m;i++)
{
if(dp[i][j]<inff)
{
for(k=p[i];k+1;k=E[k].next)
{
w=E[k].en;
zt=j;
if(((1<<(w-1))&zt)==0)
{
zt+=1<<(w-1);
if(dp[w][zt]>dp[i][j]+E[k].len)
{
dp[w][zt]=dp[i][j]+E[k].len;
}
}
}
}
}
}
int ans=inff;
for(i=0;i<=m;i++)
{
ans=min(ans,dp[i][(1<<m)-1]);
}
if(ans==inff)
ans=-1;
printf("%d\n",ans);
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<n;i++)
{
scanf("%s",tu[i]);
}
for(i=1;i<=m;i++)
{
scanf("%d%d%d%d",&u[i].x,&u[i].y,&v[i].x,&v[i].y);
u[i].x--,v[i].x--,u[i].y--,v[i].y--;
}
init();
for(i=1;i<=m;i++)
{
bfs(v[i].x,v[i].y);
add(0,i,0);
for(j=1;j<=m;j++)
{
if(i==j)
continue;
add(i,j,dis[u[j].x][u[j].y]);
}
}
solve();
}
return 0;
}