16.两岛联通

class Solution {
public:
    void dfs(vector<vector<int>>& A,int i,int j){
        if(i<0||i>=A.size()||j<0||j>=A[0].size())return;
        if(A[i][j]==1) A[i][j]=2;
        else return;
        dfs(A,i+1,j);
        dfs(A,i-1,j);
        dfs(A,i,j+1);
        dfs(A,i,j-1);
    }
    bool bfs(vector<vector<int>>& A,set<pair<int,int> >&v){
        set<pair<int,int> >vvv=set<pair<int,int> >(v);
        for(auto vv:vvv){
            if((vv.second+1<A[0].size()&&A[vv.first][vv.second+1]==2)||(vv.second-1>=0&&A[vv.first][vv.second-1]==2)||(vv.first-1>=0&&A[vv.first-1][vv.second]==2)||(vv.first+1<A.size()&&A[vv.first+1][vv.second]==2)) return 1;
            if(vv.second+1<A[0].size()&&A[vv.first][vv.second+1]==0){A[vv.first][vv.second+1]=1;v.insert(make_pair(vv.first,vv.second+1));}
            if(vv.second-1>=0&&A[vv.first][vv.second-1]==0){A[vv.first][vv.second-1]=1;v.insert(make_pair(vv.first,vv.second-1));}
            if(vv.first+1<A.size()&&A[vv.first+1][vv.second]==0){A[vv.first+1][vv.second]=1;v.insert(make_pair(vv.first+1,vv.second));}
            if(vv.first-1>=0&&A[vv.first-1][vv.second]==0){A[vv.first-1][vv.second]=1;v.insert(make_pair(vv.first-1,vv.second));}
            v.erase(vv);
        }
        return 0;
    }
    int shortestBridge(vector<vector<int>>& A) {
        int i,j,p,q;
        for( i=0;i<A.size();i++){
            for( j=0;j<A[0].size();j++){
                if(A[i][j]==0)continue;
                else {p=i;q=j;break;}
            }
        }
        dfs(A,p,q);
        set<pair<int,int> >v;
        for(i=0;i<A.size();i++){
            for(j=0;j<A[0].size();j++){
                if(A[i][j]==1){v.insert(make_pair(i,j));}
            }
        }
        int ans=0;
        while(!bfs(A,v)){ans++;}
        return ans;
    }
};

找到一个'1'后，dfs(上下左右)找出所有相连。

找出第二座岛的所有序号，放在一个set里，每次往外拓展一层，直到碰到第一座岛