BZOJ_3049_[Usaco2013 Jan]Island Travels _状压DP+BFS

BZOJ_3049_[Usaco2013 Jan]Island Travels _状压DP+BFS

Description

Farmer John has taken the cows to a vacation out on the ocean! The cows are living on N (1 <= N <= 15) islands, which are located on an R x C grid (1 <= R, C <= 50). An island is a maximal connected group of squares on the grid that are marked as 'X', where two 'X's are connected if they share a side. (Thus, two 'X's sharing a corner are not necessarily connected.) Bessie, however, is arriving late, so she is coming in with FJ by helicopter. Thus, she can first land on any of the islands she chooses. She wants to visit all the cows at least once, so she will travel between islands until she has visited all N of the islands at least once. FJ's helicopter doesn't have much fuel left, so he doesn't want to use it until the cows decide to go home. Fortunately, some of the squares in the grid are shallow water, which is denoted by 'S'. Bessie can swim through these squares in the four cardinal directions (north, east, south, west) in order to travel between the islands. She can also travel (in the four cardinal directions) between an island and shallow water, and vice versa. Find the minimum distance Bessie will have to swim in order to visit all of the islands. (The distance Bessie will have to swim is the number of distinct times she is on a square marked 'S'.) After looking at a map of the area, Bessie knows this will be possible.

  给你一张r*c的地图，有’S’,’X’,’.’三种地形，所有判定相邻与行走都是四连通的。我们设’X’为陆地，一个’X’连通块为一个岛屿，’S’为浅水，’.’为深水。刚开始你可以降落在任一一块陆地上，在陆地上可以行走，在浅水里可以游泳。并且陆地和浅水之间可以相互通行。但无论如何都不能走到深水。你现在要求通过行走和游泳使得你把所有的岛屿都经过一边。Q：你最少要经过几个浅水区？保证有解。

 

Input

* Line 1: Two space-separated integers: R and C.
* Lines 2..R+1: Line i+1 contains C characters giving row i of the grid. Deep water squares are marked as '.', island squares are marked as 'X', and shallow water squares are marked as 'S'.

Output

* Line 1: A single integer representing the minimum distance Bessie has to swim to visit all islands.

Sample Input

5 4
XX.S
.S..
SXSS
S.SX
..SX
INPUT DETAILS: There are three islands with shallow water paths connecting some of them.

Sample Output

3
OUTPUT DETAILS: Bessie can travel from the island in the top left to the one in the middle, swimming 1 unit,
and then travel from the middle island to the one in the bottom right, swimming 2 units, for a total of 3 units.

HINT

样例解释：

 5*4的地图，先走到左上角的岛屿，再向下经过1个’S’区到达中间的岛屿，再向右经过2个’S’区到达右下角的岛屿。（最优路径不一定只有一条）

---

设f[i][j]表示连通状态为i当前在j这个岛屿的最小花费。

BFS处理出两个岛屿之间有多少个浅水区。

状压DP即可。

 

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define inf 0x3f3f3f3f
#define _min(x,y) ((x)<(y)?(x):(y))
int tx[]={1,0,0,-1};
int ty[]={0,1,-1,0};
int map[55][55],dis[17][17],f[1<<15][17],idx[55][55],vis[55][55],dep[55][55],tot,Q[5050],cnt,n,m,l,r,p1,mp[1<<15];
char ch[55];
void bfs(int sx,int sy) {
    int i; tot++; cnt++;
    l=r=0;
    Q[r++]=sx; Q[r++]=sy; idx[sx][sy]=cnt;
    while(l<r) {
        int x=Q[l++],y=Q[l++]; vis[x][y]=tot;
        for(i=0;i<4;i++) {
            int dx=x+tx[i],dy=y+ty[i];
            if(dx<1||dx>n||dy<1||dy>m) continue;
            if(map[dx][dy]==0&&vis[dx][dy]!=tot) {
                vis[dx][dy]=tot; idx[dx][dy]=cnt;
                Q[r++]=dx; Q[r++]=dy;
            }
        }
    }
}
void dfs(int x,int y,int d) {
    int i;Q[r++]=x; Q[r++]=y; dep[x][y]=d;
    int tmp=idx[x][y];
    dis[p1][tmp]=_min(dis[p1][tmp],d);
    for(i=0;i<4;i++) {
        int dx=x+tx[i],dy=y+ty[i];
        if(dx>=1&&dx<=n&&dy>=1&&dy<=m) {
            if(map[dx][dy]==0&&vis[dx][dy]!=tot) {
                vis[dx][dy]=tot;
                dfs(dx,dy,d);
            }
        }
    }
}
void get_dis() {
    p1++;
    int i,j; tot++; dis[p1][p1]=0;l=r=0;
    for(i=1;i<=n;i++) for(j=1;j<=m;j++) if(idx[i][j]==p1) Q[r++]=i,Q[r++]=j,vis[i][j]=tot,dep[i][j]=0;
    while(l<r) {
        int x=Q[l++],y=Q[l++];
        for(i=0;i<4;i++) {
            int dx=x+tx[i],dy=y+ty[i];
            if(dx<1||dx>n||dy<1||dy>m) continue;
            if(map[dx][dy]==1&&vis[dx][dy]!=tot) {
                vis[dx][dy]=tot;
                dfs(dx,dy,dep[x][y]+1);
            }
        }
    }
}
int main() {
    scanf("%d%d",&n,&m);
    int i,j,k,o,q;
    for(i=1;i<=n;i++) {
        scanf("%s",ch+1);
        for(j=1;j<=m;j++) {
            if(ch[j]=='.') map[i][j]=2;
            else if(ch[j]=='X') map[i][j]=0;
            else map[i][j]=1;
        }
    }
    for(i=1;i<=n;i++) {
        for(j=1;j<=m;j++) {
            if(!vis[i][j]&&map[i][j]==0) bfs(i,j);
        }
    }
    memset(f,0x3f,sizeof(f));
    for(i=1;i<=cnt;i++) f[1<<(i-1)][i]=0,mp[1<<(i-1)]=i;
    memset(dis,0x3f,sizeof(dis));
    for(i=1;i<=cnt;i++) get_dis();
    int mask=(1<<cnt)-1;
    for(i=1;i<mask;i++) {
        for(o=i;o;o-=o&(-o)) {
            j=mp[o&(-o)];
            for(q=mask-i;q;q-=q&(-q)) {
                k=mp[q&(-q)];
                f[i|(1<<(k-1))][k]=_min(f[i|(1<<(k-1))][k],f[i][j]+dis[j][k]);
            }
        }
    }
    int ans=1<<30;
    for(i=1;i<=cnt;i++) ans=_min(ans,f[mask][i]);
    printf("%d\n",ans);
}