Codeforces 591E Three States【优先队列Bfs+思维】

E. Three States

time limit per test

5 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

The famous global economic crisis is approaching rapidly, so the states of Berman, Berance and Bertaly formed an alliance and allowed the residents of all member states to freely pass through the territory of any of them. In addition, it was decided that a road between the states should be built to guarantee so that one could any point of any country can be reached from any point of any other State.

Since roads are always expensive, the governments of the states of the newly formed alliance asked you to help them assess the costs. To do this, you have been issued a map that can be represented as a rectangle table consisting of n rows and m columns. Any cell of the map either belongs to one of three states, or is an area where it is allowed to build a road, or is an area where the construction of the road is not allowed. A cell is called passable, if it belongs to one of the states, or the road was built in this cell. From any passable cells you can move up, down, right and left, if the cell that corresponds to the movement exists and is passable.

Your task is to construct a road inside a minimum number of cells, so that it would be possible to get from any cell of any state to any cell of any other state using only passable cells.

It is guaranteed that initially it is possible to reach any cell of any state from any cell of this state, moving only along its cells. It is also guaranteed that for any state there is at least one cell that belongs to it.

Input

The first line of the input contains the dimensions of the map n and m (1 ≤ n, m ≤ 1000) — the number of rows and columns respectively.

Each of the next n lines contain m characters, describing the rows of the map. Digits from 1 to 3 represent the accessory to the corresponding state. The character '.' corresponds to the cell where it is allowed to build a road and the character '#' means no construction is allowed in this cell.

Output

Print a single integer — the minimum number of cells you need to build a road inside in order to connect all the cells of all states. If such a goal is unachievable, print -1.

Examples

input

4 511..2#..22#.323.#333

output

2

input

1 51#2#3

output

-1

题目大意：

现在我们有三个国家，每个国家都是一个联通块，用数字1、2和3来表示。

现在我们希望在点的位子上建立路，使得三个国家连通起来，现在问需要建立路的最小个数。

思路：

因为只有三个国家，所以我们用每个国家求一次单源最短路，记录dist【i】【x】【y】表示i号国家想要和点（x，y）进行连通的话，需要建立的路的个数。

那么对应，我们只要O（nm）的去枚举一个交汇点，那么dist【1】【x】【y】+dist【2】【x】【y】+dist【3】【x】【y】就是答案，如果点（x，y）是点的话，那么对应还要减去2.

我们去Bfs的时候，注意到达每个点的时候，需要求最短路，所以我们这里直接用队列Bfs的话可能会TLE，用优先队列优化一下就不会了TAT、

Ac代码：

#include<stdio.h>#include<queue>#include<string.h>#include<algorithm>#include<stack>using namespace std;struct node{    int x,y,step;    bool friend operator <(node a ,node b)    {        return a.step>b.step;    }}now,nex;char a[1050][1050];int dodo[400];int dist[5][1050][1050];int vis[1050][1050];int fx[4]={0,0,1,-1};int fy[4]={1,-1,0,0};int n,m;void Bfs(int x,int y,int dd){    memset(vis,0,sizeof(vis));    priority_queue<node>s;    for(int i=0;i<n;i++)    {        for(int j=0;j<m;j++)        {            dist[dd][i][j]=0x3f3f3f3f;            if(a[i][j]-'0'==dd)            {                now.x=i;                now.y=j;                now.step=0;                s.push(now);                dist[dd][i][j]=0;            }        }    }    while(!s.empty())    {        now=s.top();        vis[now.x][now.y]=0;        s.pop();        for(int i=0;i<4;i++)        {            nex.x=now.x+fx[i];            nex.y=now.y+fy[i];            nex.step=now.step;            if(nex.x>=0&&nex.x<n&&nex.y>=0&&nex.y<m&&a[nex.x][nex.y]!='#')            {                if(a[nex.x][nex.y]=='.')                {                    nex.step++;                }                if(dist[dd][nex.x][nex.y]>nex.step)                {                    dist[dd][nex.x][nex.y]=nex.step;                    if(vis[nex.x][nex.y]==0)                    {                        s.push(nex);                    }                }            }        }    }}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(dodo,0,sizeof(dodo));        for(int i=0;i<n;i++)scanf("%s",a[i]);        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                if(dodo[a[i][j]]==0&&a[i][j]>='1'&&a[i][j]<='3')                {                    Bfs(i,j,a[i][j]-'0');dodo[a[i][j]]=1;                }            }        }        int output=0x3f3f3f3f;        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                if(a[i][j]!='#')                {                    int flag=0;                    int ans=0;                    for(int dd=1;dd<=3;dd++)                    {                        if(dist[dd][i][j]==0x3f3f3f3f)flag=1;                        ans+=dist[dd][i][j];                    }                    if(flag==1)continue;                    if(a[i][j]=='.')ans-=2;                    output=min(output,ans);                }            }        }        if(output==0x3f3f3f3f)printf("-1\n");        else        printf("%d\n",output);    }}