题目链接:点击打开链接
题意:
给定n*m的矩阵,可以在矩阵中画出题目图片里黑色的线条。
任选一个点,左下右上的顺序走,走出图里的情况。使得走过的数字和最大,问最大的和。
思路:
题解比较详细了
import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.PrintWriter; import java.math.BigInteger; import java.text.DecimalFormat; import java.util.ArrayDeque; import java.util.ArrayList; import java.util.Arrays; import java.util.Collection; import java.util.Collections; import java.util.Comparator; import java.util.Deque; import java.util.HashMap; import java.util.Iterator; import java.util.LinkedList; import java.util.Map; import java.util.PriorityQueue; import java.util.Scanner; import java.util.Stack; import java.util.StringTokenizer; import java.util.TreeMap; import java.util.TreeSet; import java.util.Queue; import java.io.File; import java.io.FileInputStream; import java.io.FileNotFoundException; import java.io.FileOutputStream; public class Main { int n, m; int[][] a = new int[N][N], sum = new int[N][N]; int[][][] s; void work() throws Exception { n = Int(); m = Int(); for(int i = 1; i<= n; i++) for(int j = 1; j <= m; j++){ sum[i][j] = a[i][j] = Int(); if(i>0) sum[i][j] += sum[i-1][j]; if(j>0) sum[i][j] += sum[i][j-1]; if(i>0&&j>0) sum[i][j] -= sum[i-1][j-1]; } s = new int[n+1][][]; for(int i = 1; i <= n; i++)s[i] = new int[m+1][]; for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++){ s[i][j] = new int[min(n+2-i,m+2-j)/2]; s[i][j][0] = a[i][j]; } int ans = -inf; for(int k = 1; k <= 250; k++) for(int i = 1; i+2*k <= n; i++) for(int j = 1; j + 2*k <= m; j++) { s[i][j][k] = sum[i+2*k][j+2*k]-sum[i-1][j+2*k]-sum[i+2*k][j-1]+sum[i-1][j-1]-s[i+1][j+1][k-1]-a[i+1][j]; ans = max(ans, s[i][j][k]); } out.println(ans); } public static void main(String[] args) throws Exception { Main wo = new Main(); in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); // in = new BufferedReader(new InputStreamReader(new FileInputStream(new // File("input.txt")))); // out = new PrintWriter(new File("output.txt")); wo.work(); out.close(); } static int N = 505; static int M = N*2; DecimalFormat df = new DecimalFormat("0.0000"); static int inf = (int) 1e9; static long inf64 = (long) 1e18; static double eps = 1e-8; static double Pi = Math.PI; static int mod = (int) 1e9 + 7; private String Next() throws Exception { while (str == null || !str.hasMoreElements()) str = new StringTokenizer(in.readLine()); return str.nextToken(); } private int Int() throws Exception { return Integer.parseInt(Next()); } private long Long() throws Exception { return Long.parseLong(Next()); } private double Double() throws Exception { return Double.parseDouble(Next()); } StringTokenizer str; static Scanner cin = new Scanner(System.in); static BufferedReader in; static PrintWriter out; /* class Edge{ int from, to, dis, nex; Edge(){} Edge(int from, int to, int dis, int nex){ this.from = from; this.to = to; this.dis = dis; this.nex = nex; } } Edge[] edge = new Edge[M<<1]; int[] head = new int[N]; int edgenum; void init_edge(){ for(int i = 0; i < N; i++)head[i] = -1; edgenum = 0;} void add(int u, int v, int dis){ edge[edgenum] = new Edge(u, v, dis, head[u]); head[u] = edgenum++; }/**/ int upper_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A; int pos = r; r--; while (l <= r) { int mid = (l + r) >> 1; if (A[mid] <= val) { l = mid + 1; } else { pos = mid; r = mid - 1; } } return pos; } int lower_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A; int pos = r; r--; while (l <= r) { int mid = (l + r) >> 1; if (A[mid] < val) { l = mid + 1; } else { pos = mid; r = mid - 1; } } return pos; } int Pow(int x, int y) { int ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; y >>= 1; x = x * x; } return ans; } double Pow(double x, int y) { double ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; y >>= 1; x = x * x; } return ans; } int Pow_Mod(int x, int y, int mod) { int ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; ans %= mod; y >>= 1; x = x * x; x %= mod; } return ans; } long Pow(long x, long y) { long ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; y >>= 1; x = x * x; } return ans; } long Pow_Mod(long x, long y, long mod) { long ans = 1; while (y > 0) { if ((y & 1) > 0) ans *= x; ans %= mod; y >>= 1; x = x * x; x %= mod; } return ans; } int gcd(int x, int y) { if (x > y) { int tmp = x; x = y; y = tmp; } while (x > 0) { y %= x; int tmp = x; x = y; y = tmp; } return y; } int max(int x, int y) { return x > y ? x : y; } int min(int x, int y) { return x < y ? x : y; } double max(double x, double y) { return x > y ? x : y; } double min(double x, double y) { return x < y ? x : y; } long max(long x, long y) { return x > y ? x : y; } long min(long x, long y) { return x < y ? x : y; } int abs(int x) { return x > 0 ? x : -x; } double abs(double x) { return x > 0 ? x : -x; } long abs(long x) { return x > 0 ? x : -x; } boolean zero(double x) { return abs(x) < eps; } double sin(double x) { return Math.sin(x); } double cos(double x) { return Math.cos(x); } double tan(double x) { return Math.tan(x); } double sqrt(double x) { return Math.sqrt(x); } }
时间: 2024-10-27 05:13:44