poj2019(二维线段树)

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大致题意：给定一个n*n矩阵以及k个查询。每个查询中给出两个数x,y,则该查询询问以第x行第y列为左上角的b*b

矩阵中元素最大值与最小值之差。

思路：二维问题简单化，可先考虑一维的情形，即考虑区间内元素最大最小值问题。这个可以用RMQ解决，当然，

写线段树也可以，这道题我就直接以线段树为基础了。主要的式子就是maxval[o]=max(maxval[o*2],maxval[o*2+1])，

然后二维相对一维，差不多就数组多了一维，参数，边界等稍微复杂点。这道就当成自己二维线段树模板题啦~~~

代码如下：

import java.io.BufferedReader;import java.io.IOException;import java.io.InputStream;import java.io.InputStreamReader;import java.util.StringTokenizer;class Reader {static BufferedReader reader;static StringTokenizer tokenizer;static void init(InputStream input) {reader = new BufferedReader(new InputStreamReader(input));tokenizer = new StringTokenizer("");}static String next() throws IOException {while (!tokenizer.hasMoreTokens()) {tokenizer = new StringTokenizer(reader.readLine());}return tokenizer.nextToken();}static int nextInt() throws IOException {return Integer.parseInt(next());}}class Num {int max, min;public Num(int max, int min) {super();this.max = max;this.min = min;}public Num() {super();// TODO Auto-generated constructor stub}}public class Main {/** * @param args */static int n, b, k, x, y;static int val[][];static int max[][], min[][];private static void btree1(int o, int l, int r) {if (l == r) {btree2(o, l, r, 1, 1, n);return;}int mid = (l + r) / 2;btree1(o * 2, l, mid);btree1(o * 2 + 1, mid + 1, r);btree2(o, l, r, 1, 1, n);}private static int getmax(int a, int b) {return a > b ? a : b;}private static int getmin(int a, int b) {return a < b ? a : b;}private static void btree2(int xo, int xl, int xr, int o, int l, int r) {if (l == r) {if (xl == xr) {max[xo][o] = val[xl][l];min[xo][o] = val[xr][l];} else {max[xo][o] = getmax(max[xo * 2][o], max[xo * 2 + 1][o]);min[xo][o] = getmin(min[xo * 2][o], min[xo * 2 + 1][o]);}return;}int mid = (l + r) / 2;btree2(xo, xl, xr, o * 2, l, mid);btree2(xo, xl, xr, o * 2 + 1, mid + 1, r);max[xo][o] = getmax(max[xo][o * 2], max[xo][o * 2 + 1]);min[xo][o] = getmin(min[xo][o * 2], min[xo][o * 2 + 1]);}private static Num query1(int o, int l, int r) {if ((l >= x) && (r <= x + b - 1))return query2(o, l, r, 1, 1, n);int mid = (l + r) / 2;Num num1, num2;num1 = new Num(0, Integer.MAX_VALUE);if (mid >= x)num1 = query1(o * 2, l, mid);if (mid < x + b - 1) {num2 = query1(o * 2 + 1, mid + 1, r);num1.max = getmax(num1.max, num2.max);num1.min = getmin(num1.min, num2.min);}return num1;}private static Num query2(int xo, int xl, int xr, int o, int l, int r) {if ((l >= y) && (r <= y + b - 1)) {return new Num(max[xo][o], min[xo][o]);}int mid = (l + r) / 2;Num num1, num2;num1 = new Num(0, Integer.MAX_VALUE);if (mid >= y)num1 = query2(xo, xl, xr, o * 2, l, mid);if (mid < y + b - 1) {num2 = query2(xo, xl, xr, o * 2 + 1, mid + 1, r);num1.max = getmax(num1.max, num2.max);num1.min = getmin(num1.min, num2.min);}return num1;}public static void main(String[] args) throws IOException {// TODO Auto-generated method stubReader.init(System.in);n = Reader.nextInt();b = Reader.nextInt();k = Reader.nextInt();val = new int[n + 1][n + 1];for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)val[i][j] = Reader.nextInt();max = new int[4 * n + 1][4 * n + 1];min = new int[4 * n + 1][4 * n + 1];btree1(1, 1, n);for (int i = 1; i <= k; i++) {x = Reader.nextInt();y = Reader.nextInt();System.out.println(query1(1, 1, n).max - query1(1, 1, n).min);}}}