WebService之CXF註解之三（Service介面實作類別）

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Paint the Grid ReloadedTime Limit: 2 Seconds      Memory Limit: 65536 KB

Leo has a grid with N rows and M columns. All cells are painted with either black or white initially.

Two cells A and B are called connected if they share an edge and they are in the same color, or there exists a cellC connected to both A and B.

Leo wants to paint the grid with the same color. He can make it done in multiple steps. At each step Leo can choose a cell and flip the color (from black to white or from white to black) of all cells connected to it. Leo wants to know the minimum number of steps he needs to make all cells in the same color.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers N and M (1 <= N, M <= 40). Then N lines follow. Each line contains a string withN characters. Each character is either ‘X‘ (black) or ‘O‘ (white) indicates the initial color of the cells.

Output

For each test case, output the minimum steps needed to make all cells in the same color.

Sample Input

22 2OXOX3 3XOXOXOXOX

Sample Output

12

Hint

For the second sample, one optimal solution is:

Step 1. flip (2, 2)

XOXOOOXOX

Step 2. flip (1, 2)

XXXXXXXXX

題意：

給你一個染色的 N×M 的格子，你每次可以選擇一個連通塊，將其顏色取反。問最後將整個格子變為同色的最小步數。

思路：

其實就是一個最短路，將連通塊縮點然後不同種的連通塊之間連邊，那麼將整個格子變為同色的最小步數就等於一個點到地圖上最遠點的距離了。

代碼：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:1024000000,1024000000")#define maxn 45#define MAXN 2700005#define OO (1<<31)-1#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int n,m,ans,cnt,lev;char mp[maxn][maxn];int num[maxn][maxn];int dx[]={0,0,-1,1};int dy[]={-1,1,0,0};bool vis[maxn][maxn],app[1605][1605];int p[1700];struct Node{    int v;    int next;}edge[MAXN];void addedge(int u,int v){    cnt++;    edge[cnt].v=v;    edge[cnt].next=p[u];    p[u]=cnt;}bool isok(int x,int y){    if(x<1||x>n||y<1||y>m) return false ;    return true ;}void dfs(int x,int y){    num[x][y]=lev;    int nx,ny,i;    for(i=0;i<4;i++)    {        nx=x+dx[i]; ny=y+dy[i];        if(isok(nx,ny)&&!vis[nx][ny]&&mp[nx][ny]==mp[x][y])        {            vis[nx][ny]=1;            dfs(nx,ny);        }    }}void presolve(){    memset(p,0,sizeof(p));    int i,j,k,t,nx,ny;    lev=0;    memset(vis,0,sizeof(vis));    for(i=1;i<=n;i++)    {        for(j=1;j<=m;j++)        {            if(!vis[i][j])            {                lev++;                vis[i][j]=1;                dfs(i,j);            }        }    }    cnt=0;    memset(app,0,sizeof(app));    for(i=1;i<=n;i++)    {        for(j=1;j<=m;j++)        {            for(k=0;k<4;k++)            {                nx=i+dx[k]; ny=j+dy[k];                if(isok(nx,ny)&&num[nx][ny]!=num[i][j]&&!app[num[nx][ny]][num[i][j]])                {                    app[num[nx][ny]][num[i][j]]=1;                    addedge(num[nx][ny],num[i][j]);                }            }        }    }}struct fuck{    int dis;    int num;}t,f;bool VVV[2700];queue<fuck> Q;int bfs(int now){    memset(VVV,0,sizeof(VVV));    t.dis=0;    t.num=now;    VVV[now]=1;    while(!Q.empty()) Q.pop();    Q.push(t);    int maxs=0;    while(!Q.empty())    {        t=Q.front();Q.pop();        if(t.dis>ans) return INF;        for(int i=p[t.num];i!=0;i=edge[i].next)        {            int to=edge[i].v;            if(!VVV[to])            {                VVV[to]=1;                f.num=to;                f.dis=t.dis+1;                maxs=max(maxs,f.dis);                Q.push(f);            }        }    }    return maxs;}int main(){    int i,j,t,u,v,w;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(i=1;i<=n;i++)        {            scanf("%s",mp[i]+1);        }        presolve();        ans=INF;        for(int i=1;i<=lev;i++)        {            ans=min(ans,bfs(i));        }        printf("%d\n",ans);    }    return 0;}