題干：

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 cell?C?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?Tindicating 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 with?N?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

2 2 2 OX OX 3 3 XOX OXO XOX

Sample Output

1 2

Hint

For the second sample, one optimal solution is:

Step 1. flip (2, 2)

XOX OOO XOX

Step 2. flip (1, 2)

XXX XXX XXX

題目大意：

字符一樣并且相鄰的即為連通。每次可翻轉一個連通塊X(O)的顏色，問至少改變幾次使得圖上所有字符都相等

解題報告：

思路1：直接枚舉起點，然后暴力搜。這個步驟搜的O那下一個步驟就搜X。。

思路2：dfs連通塊縮點，枚舉起點bfs求最短路，注意要用連通塊來求bfs，，不然會T。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 55; char s[MAX][MAX]; int bk[MAX][MAX]; int nx[4] = {1,0,-1,0}; int ny[4] = {0,1,0,-1}; int tot,n,m; int vis[MAX*MAX]; vector<int> vv[MAX*MAX]; struct Node {int id;int t;Node(){}Node(int id,int t):id(id),t(t){} }; bool ok(int x,int y) {if(x>=1&&x<=n&&y>=1&&y<=m) return 1 ;else return 0 ; } void dfs(int x,int y,char tar) {bk[x][y] = tot; for(int k = 0; k<4; k++) {int tx = x + nx[k];int ty = y + ny[k];if(ok(tx,ty) == 0) continue;if(s[tx][ty] == tar) {if(bk[tx][ty] == -1) dfs(tx,ty,tar);}else if(bk[tx][ty] != -1) {vv[tot].pb(bk[tx][ty]);vv[bk[tx][ty]].pb(tot);}} }int bfs(int x) {memset(vis,0,sizeof vis);queue<Node> q;int res = 0;q.push(Node(x,0));vis[x]=1;while(q.size()) {Node cur = q.front();q.pop();int col = cur.id;res = cur.t;int up = vv[col].size();for(int i = 0; i<up; i++) {int v = vv[col][i];if(vis[v]) continue;vis[v]=1;q.push(Node(v,cur.t+1));} } return res; } int main() {int t;cin>>t;while(t--) {tot=0;memset(bk,-1,sizeof bk);scanf("%d%d",&n,&m);for(int i = 1; i<=n; i++) {scanf("%s",s[i]+1);}for(int i = 1; i<=n*m; i++) vv[i].clear();for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(bk[i][j]== -1) {tot++;dfs(i,j,s[i][j]);}}}int ans = 9999999;for(int i = 1; i<=tot; i++) ans = min(ans,bfs(i));printf("%d\n",ans);}return 0 ; }

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