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827.cpp
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827.cpp
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__________________________________________________________________________________________________
sample 8 ms submission
class Solution {
public:
int a[50][50]={0};//对岛分类编号
int area[700]={0};//不同编号的岛对应不同的面积
int islandnum=0;//编号
void search(vector<vector<int>>& grid,int x,int y)
{
int X=grid.size();
int Y=grid[0].size();
if(grid[x][y]==1&&a[x][y]==0)
{
area[islandnum]++;//累计相同编号的岛的陆地
a[x][y]=islandnum;//给陆地附上编号
if(x-1>=0) search(grid,x-1,y);//四个方向
if(x+1<X) search(grid,x+1,y);
if(y-1>=0) search(grid,x,y-1);
if(y+1<Y) search(grid,x,y+1);
}
}
int largestIsland(vector<vector<int>>& grid) {
int X=grid.size();
int Y=grid[0].size();
for(int i=0;i<X;i++)
{
for(int j=0;j<Y;j++)
{
if(grid[i][j]==1&&a[i][j]==0)
{
islandnum++;//找到的第一个岛,应该从编号1开始
search(grid,i,j);
}
}
}
int b[4]={0};//海的四个方向的岛的编号,上下左右
int maxsum=0,sum;
for(int i=0;i<X;i++)
{
for(int j=0;j<Y;j++)
{
if(grid[i][j]==0)
{
sum=1;//毕竟填海嘛,面积得从1算起
if(i-1<0) b[0]=0;
else b[0]=a[i-1][j];//上没有越界,则附值上方的岛的编号
if(i+1>=X) b[1]=0;
else b[1]=a[i+1][j];//同理
if(j-1<0) b[2]=0;
else b[2]=a[i][j-1];//同理
if(j+1>=Y) b[3]=0;
else b[3]=a[i][j+1];//同理
for(int k=0;k<4;k++)
{
if(b[k]!=0)
{
sum+=area[b[k]];//加上该方向的岛的编号的面积
for(int l=k+1;l<4;l++)
{
if(b[l]==b[k]) b[l]=0;//保证四个方向不会出现相同编号的岛
}
//b[k]=0;
}
if(sum>maxsum) maxsum=sum;
}
}
}
}
if(maxsum==0) maxsum=X*Y;//没有海的情况下
return maxsum;
}
};
__________________________________________________________________________________________________
sample 12 ms submission
// with changes as in a submission
class Solution {
vector<vector<int> > dir = { { -1, 0 }, { 0, -1 }, { 1, 0 }, { 0, 1 } };
public:
int largestIsland(vector<vector<int> >& grid) {
int index = 2, m = grid.size(), n = grid[0].size();
vector<int> area(2);
int res = 0;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (grid[i][j] == 1) {
int a = dfs(i, j, grid, index++);
res = max(res, a);
area.push_back(a);
}
vector<int> seen;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (grid[i][j] == 0) {
//unordered_set<int> hs;
int r = 1;
for (int d = 0; d < 4; d++) {
int ii = i + dir[d][0], jj = j + dir[d][1];
if (ii >= 0 && ii < m && jj >= 0 && jj < n && area[grid[ii][jj]] > 0) { // !hs.count(grid[ii][jj])) {
int& a = area[grid[ii][jj]];
r += a;
a *= -1;
seen.push_back(grid[ii][jj]); // to restore later
//hs.insert(grid[ii][jj]);
}
}
res = max(res, r);
for (int x : seen) area[x] *= -1; // restore area values
seen.clear();
}
return res;
}
int dfs(int i, int j, vector<vector<int> >& g, int x) {
g[i][j] = x;
int res = 1;
for (int d = 0; d < 4; d++) {
int ii = i + dir[d][0], jj = j + dir[d][1];
if (ii >= 0 && ii < g.size() && jj >= 0 && jj < g[0].size() && g[ii][jj] == 1)
res += dfs(ii, jj, g, x);
}
return res;
}
};
__________________________________________________________________________________________________
sample 16 ms submission
class UnionFind {
public:
UnionFind(int n): sz(n,0),id(n,0) {
for (int i=0;i<n;i++) id[i]=i;
}
bool find(int p, int q) {
if (root(p)!=root(q)) return false;
else return true;
}
void merge(int p, int q) {
int rp=root(p), rq=root(q);
if (rp==rq) return;
if (sz[rp]>=sz[rq]) {
id[rq]=rp;
sz[rp]+=sz[rq];
}
else {
id[rp]=rq;
sz[rq]+=sz[rp];
}
}
int root(int p) {
while (id[p]!=p) {
id[p]=id[id[p]];
p=id[p];
}
return p;
}
int trial(int x, int y, int n, int m) {
int index=x*n+y;
vector<int> r(4,-1),s(4,0);
if (x>0) {r[0]=root(index-m); s[0]=sz[r[0]];}//up
if (y>0) {r[1]=root(index-1); s[1]=sz[r[1]];}//left
if (x<n-1) {r[2]=root(index+m); s[2]=sz[r[2]];}//below
if (y<m-1) {r[3]=root(index+1); s[3]=sz[r[3]];}//right
//cout<<x<<","<<y<<":"<<s[0]<<" "<<s[1]<<" "<<s[2]<<" "<<s[3]<<endl;
for (int i=0;i<4;i++) {
if (r[i]==-1) continue;
for (int j=i+1;j<4;j++) {
if (r[i]==r[j]) s[i]=0;
}
}
return s[0]+s[1]+s[2]+s[3]+1;
}
void add(int p) {
sz[p]=1;
}
//private:
vector<int> id;
vector<int> sz;
};
class Solution {
public:
int largestIsland(vector<vector<int>>& grid) {
int n=grid.size(),m=grid[0].size();
UnionFind obj(n*m);
for (int i=0;i<n;i++)
for (int j=0;j<m;j++) {
int index=i*m+j;
if (grid[i][j]==1) {
obj.add(index);
if (i>0 && grid[i-1][j]==1) obj.merge(index,index-m);
if (j>0 && grid[i][j-1]==1) obj.merge(index,index-1);
}
}
//cout<<"1111"<<endl;
int res=0;
for (int i=0;i<n;i++)
for (int j=0;j<m;j++) {
if (grid[i][j]==0) {
res=max(res,obj.trial(i,j,n,m));
}
}
if (res==0) res=n*m;
return res;
}
};
__________________________________________________________________________________________________
sample 20 ms submission
class Solution {
public:
int largestIsland(vector<vector<int>>& grid) {
if(grid.size() == 0 || grid[0].size() == 0) return 0;
int rows = grid.size(), cols = grid[0].size();
vector<int> areas;
vector<vector<int>> ids(rows, vector<int>(cols, -1));
for(int i = 0; i < rows; i++){
for(int j = 0; j < cols; j++){
if(grid[i][j] == 1 && ids[i][j] < 0){
int k = areas.size();
areas.push_back(dfs(grid, i, j, ids, k));
}
}
}
int res = 0;
int dirs[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
for(int i = 0; i < rows; i++){
for(int j = 0; j < cols; j++){
if(grid[i][j] == 0){
set<int> s;
int cur = 1;
for(int k = 0; k < 4; k++){
int x = i + dirs[k][0], y = j + dirs[k][1];
if(x >= 0 && x < grid.size() && y >= 0 && y < grid[0].size() && grid[x][y] == 1 && s.count(ids[x][y]) == 0){
s.insert(ids[x][y]);
cur += areas[ids[x][y]];
}
}
res = max(res, cur);
}
}
}
return res == 0 ? rows*cols : res;
}
int dfs(const vector<vector<int>>& grid, int row, int col, vector<vector<int>>& gridIds, int id){
int dirs[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
int area = 1;
gridIds[row][col] = id;
for(int i = 0; i < 4; i++){
int x = row + dirs[i][0], y = col + dirs[i][1];
if(x >= 0 && x < grid.size() && y >= 0 && y < grid[0].size() && grid[x][y] == 1 && gridIds[x][y] < 0) {
area += dfs(grid, x, y, gridIds, id);
}
}
return area;
}
};
__________________________________________________________________________________________________
sample 10004 kb submission
class Solution {
int color = 2;
int count = 0;
unordered_map<int, int> m;
int curmax = 0;
public:
void dfs(vector<vector<int>> &grid, int i, int j) {
if (grid[i][j] != 1) return;
grid [i][j] = color;
count++;
if (i > 0) dfs(grid, i-1, j);
if (i < grid.size()-1) dfs(grid, i+1, j);
if (j > 0) dfs(grid, i, j-1);
if (j < grid[0].size()-1) dfs(grid, i, j+1);
}
int largestIsland(vector<vector<int>>& grid) {
// color the grid
for (int i = 0; i < grid.size(); i++) {
for (int j = 0; j < grid[0].size(); j++) {
if (grid[i][j] == 1) {
dfs(grid, i, j);
m[color] = count;
if (count > curmax) {
curmax = count;
}
count = 0;
color++;
}
}
}
for (int i = 0; i < grid.size(); i++) {
for (int j = 0; j < grid[0].size(); j++) {
if (grid[i][j] == 0) {
int a = 0;
int b = 0;
int c = 0;
int d = 0;
int temp = 1;
if (i > 0) {
a = grid[i-1][j];
}
if (i < grid.size()-1) {
b = grid[i+1][j];
}
if (j > 0) {
c = grid[i][j-1];
}
if (j < grid[0].size()-1) {
d = grid[i][j+1];
}
if (a != 0) {
temp += m[a];
}
if (b != 0 && b != a) {
temp += m[b];
}
if (c != 0 && c != b && c != a) {
temp += m[c];
}
if (d != 0 && d != c && d != b && d != a) {
temp += m[d];
}
if (temp > curmax) {
curmax = temp;
}
}
}
}
return curmax;
}
};
__________________________________________________________________________________________________