【Codeforces 375B】Maximum Submatrix 2
【链接】 我是链接,点我呀:)
【题意】
【题解】
设cnt[i][j] 表示(i,j)这个点往右连续的1的个数 我们枚举列j 然后对于第j列的cnt[1..n][j] 我们把这n个数字排个序(升序) 然后顺序枚举这n个数字 假设我们枚举到了第i个数字,显然第i~n这n-i+1个数字是能组成一个宽为cnt[i][j]长为n-i+1的矩形的 且这些矩形的左边界都是i 这就是这道题的技巧所在了 找到最短的宽度,让比它长的都和它适应 秒啊【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static int N = (int)5000;
static class Task{
int n,m;
String s[] = new String[N+10];
int cnt[][] = new int[N+10][N+10];
int col[][] = new int[N+10][N+10];
public void solve(InputReader in,PrintWriter out) {
n = in.nextInt();m = in.nextInt();
for (int i = 1;i <= n;i++) {
s[i] = in.next();
StringBuilder sb = new StringBuilder(s[i]);
sb.insert(0, ' ');
s[i] = sb.toString();
}
for (int i = 1;i <= n;i++) {
for (int j = m;j >= 1;j--) {
if (s[i].charAt(j)=='0') {
cnt[i][j] = 0;
}else {
cnt[i][j] = cnt[i][j+1]+1;
}
}
}
for (int j = m;j >= 1;j--) {
for (int i = 1;i <= n;i++)
{
col[j][i] = cnt[i][j];
}
}
long ans = 0;
for (int i = 1;i <= m;i++) {
Arrays.sort(col[i],1,n+1);
for (int j = 1;j <= n;j++) {
ans = Math.max(ans,col[i][j]*(n-j+1));
}
}
out.println(ans);
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}