Red Huang

Red Huang

uva 10944

TSP 也就是 Hamilton Path 的最小成本

//  
//        GGGGGGGGGGGGG        CCCCCCCCCCCCC               AAA  
//     GGG::::::::::::G     CCC::::::::::::C              A:::A  
//   GG:::::::::::::::G   CC:::::::::::::::C             A:::::A  
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C            A:::::::A  
// G:::::G       GGGGGG C:::::C       CCCCCC           A:::::::::A  
//G:::::G              C:::::C                        A:::::A:::::A  
//G:::::G              C:::::C                       A:::::A A:::::A  
//G:::::G    GGGGGGGGGGC:::::C                      A:::::A   A:::::A  
//G:::::G    G::::::::GC:::::C                     A:::::A     A:::::A  
//G:::::G    GGGGG::::GC:::::C                    A:::::AAAAAAAAA:::::A  
//G:::::G        G::::GC:::::C                   A:::::::::::::::::::::A  
// G:::::G       G::::G C:::::C       CCCCCC    A:::::AAAAAAAAAAAAA:::::A  
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C   A:::::A             A:::::A  
//   GG:::::::::::::::G   CC:::::::::::::::C  A:::::A               A:::::A  
//     GGG::::::GGG:::G     CCC::::::::::::C A:::::A                 A:::::A  
//        GGGGGG   GGGG        CCCCCCCCCCCCCAAAAAAA                   AAAAAAA  
#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <cmath>  
#include <climits>  
#include <vector>  
#include <set>  
#include <map>  
#include <queue>  
#include <cctype>  
#include <utility>  
#include <ctime>  
using namespace std;  
#ifdef DEBUG  
#define VAR(a,b) decltype(b) a=(b)  
#define debug(...) printf("DEBUG: "),printf(__VA_ARGS__)  
#define gettime() end_time=clock();printf("now running time is %.7f\\n",(float)(end_time - start_time)/CLOCKS_PER_SEC);  
#else  
#define VAR(a,b) __typeof(b) a=(b)  
#define debug(...)  
#define gettime()  
#endif  
typedef unsigned int uint;  
typedef long long int Int;  
typedef unsigned long long int UInt;  
#define Set(a,s) memset(a,s,sizeof(a))  
#define Write(w) freopen(w,"w",stdout)  
#define Read(r) freopen(r,"r",stdin)  
#define Pln() printf("\\n")  
#define I_de(x,n)for(int i=0;i<n;i++)printf("%d ",x[i]);Pln()  
#define De(x)printf(#x"%d\\n",x)  
#define For(i,x)for(int i=0;i<x;i++)  
#define CON(x,y) x##y  
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\  
    printf("%d ",dp[htx][hty]);}Pln();}  
#define M 30  
#define PII pair<int,int>  
#define PB push_back  
#define oo INT_MAX  
#define Set_oo 0x3f  
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)  
#define eps 1e-6  
#define X first  
#define Y second  
clock_t start_time=clock(), end_time;  
bool xdy(double x,double y){return x>y+eps;}  
bool xddy(double x,double y){return x>y-eps;}  
bool xcy(double x,double y){return x<y-eps;}  
bool xcdy(double x,double y){return x<y+eps;}  
int min3(int x,int y,int z){  
    int tmp=min(x,y);  
    return min(tmp,z);  
}  
int max3(int x,int y,int z){  
    int tmp=max(x,y);  
    return max(tmp,z);  
}  
int nx,ny;  
int nuts;  
int nutx[M],nuty[M];  
char str[M];  
int mz[M][M];  
int dp[1<<18][18];  
int dfs(int state,int x){  
    if(dp[state][x]!=-1)return dp[state][x];  
    if(state==(1<<(nuts))-1){  
        return dp[state][x]=mz[x][0];  
    }  
    int minans=oo;  
    for(int i=1;i<nuts;i++){  
        if(!((state>>i)&1)&&i!=x){  
            int h=dfs(state|(1<<i),i);  
            if(h){  
                minans=min(minans,mz[x][i]+h);  
            }  
        }  
    }  
    if(minans==oo){  
        minans=0;  
    }  
    return dp[state][x]=minans;  
  
}  
void solve(){  
    Set(mz,-1);  
    Set(dp,-1);  
    for(int i=0;i<nuts;i++){  
        for(int j=i;j<nuts;j++){  
            mz[i][j]=mz[j][i]=max(abs(nutx[i]-nutx[j]),abs(nuty[i]-nuty[j]));  
        }  
    }  
    printf("%d\\n",dfs(1,0));  
}  
int main() {  
    ios_base::sync_with_stdio(0);  
    while(~scanf("%d%d",&ny,&nx)){  
        nuts=1;  
        for(int i=0;i<ny;i++){  
            scanf("%s",str);  
            for(int j=0;j<nx;j++){  
                if(str[j]=='L')nutx[0]=j,nuty[0]=i;  
                if(str[j]=='#')nutx[nuts]=j,nuty[nuts++]=i;  
            }  
        }  
        solve();  
    }  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
}  

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