山东acm省赛（第六届）Stars

Stars

这里用尺取法

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

public class A {
	static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

	public static int nextInt() throws IOException {
		in.nextToken();
		return (int) in.nval;
	}

	public static String next() throws IOException {
		in.nextToken();
		return (String) in.sval;
	}
	
	static int v[] = new int[405]; // 获取一列的星星数量
	static int n, k;
    public static void main(String[] args) throws Exception {
        int T = nextInt();
        while(T-->0) {
        	n = nextInt();
        	k = nextInt();
        	int a[][] = new int[n + 1][n + 1];
        	int x, y;
        	for(int i = 0; i < n; i++) {
        		x = nextInt();
        		y = nextInt();
        		a[x][y] = 1;
        	}
        	
        	for(int i = 1; i <= n; i++) {
        		for(int j = 1; j <= n; j++) {
        			a[i][j] += a[i][j - 1];  // 左边的积累到右边去、
        		}
        	}
        	
        	for(int i = 1; i <= n; i++) {
        		for(int j = 1; j <= n; j++) {
        			a[i][j] += a[i - 1][j];  // 上面的积累到下面去、、
        		}
        	}
        	
        	int ans = n * n;
        	// 两额循环，遍历所有的行（选两行）
        	for(int i = 0; i <= n; i++) {
        		for(int j = i + 1; j <= n; j++) {
        			ans = Math.min(ans, (j - i) * cal(a, i, j));
        		}
        	}
        	out.println(ans);
        	out.flush();
        }
    }
    

	public static int cal(int arr[][], int l, int r) { // 数组，行坐标，以及最少选几个星星、
    	int start = 0, end = 0;  // 起点终点， （尺取法）
    	
    	for(int i = 0; i <= n; i++) {
    		v[i] = arr[r][i] - arr[l][i];  // 获取一列的星星数量、、
    	}
    	
    	while(end <= n && v[end] < k) {
    		end++;
    	}
    	
    	if(end > n) {
    		return 0x3f3f3f; // 返回一个很大的值，保证这个结果不会被选择
    	}
    	
    	int ret = end;
    	while(end <= n) { // 得出来的就是最小的范围，且包含了至少k个星星
    		if(v[end] - v[start] >= k) {
    			ret = Math.min(ret, end - start++); // 满足的话，尾巴的就++
    		}else {
    			end++; // 不满足头++
    		}
    	}  
    	return ret;
    }
    
}