CodeForces 514D R2D2 and Droid Army RMQ+二分

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题目链接：点击打开链接

题意：给定n m k

下面是n*m的矩阵

最多可以操作k次，每次操作可以使任意一列上所有的数 -= 1，（ 0还是0）

要求得到连续最多的行数（每行里的整数都为0），输出任意一个方案（在每一列上操作的次数）

思路:

把每列单独考虑

枚举每行，二分找这行往下最多能清空的行数，

RMQ维护一列的最大值。

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, k;
	class RMQ{
		int[][] d = new int[N*2][20];
		int[] A = new int[N];
		int n;
		void RMQ_init() {  
		    for (int i = 1; i <= n; ++i)  
		        d[i][0] = A[i];  
		    for (int j = 1; (1 << j) <= n; ++j)  
		        for (int i = 1; i + j - 1 <= n; ++i) {  
		            d[i][j] = max(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);  
		        }  
		}  
		int RMQ(int L, int R) {  
		    int k = 0;  
		    while ((1 << (k + 1)) <= R - L + 1)  
		        ++k;  
		    return max(d[L][k], d[R - (1 << k) + 1][k]);  
		}  
	}
	RMQ[] x = new RMQ[6];
	int[] aaa = new int[6];
	void work() throws Exception{
		n = Int(); m = Int(); k = Int();
		for(int i = 1; i <= m; i++){
			x[i] = new RMQ();
		}
		for(int i = 1; i <= n; i++){
			for(int j = 1; j <= m; j++)
				x[j].A[i] = Int();
		}
		for(int i = 1; i <= m; i++){
			x[i].n = n;
			x[i].RMQ_init();
		}
		int ans = 0;
		for(int i = 1; i <= n; i++){
			int l = i, r = n;
			while(l<=r){
				int mid = (l+r)>>1;
				int sum = 0;
				for(int j = 1; j <= m; j++)
					sum += x[j].RMQ(i, mid);
				if(sum<=k){
					l = mid+1;
					if(ans<mid-i+1){
						ans=mid-i+1;
						for(int j = 1; j <= m; j++)aaa[j] = x[j].RMQ(i, mid);
					}
				}
				else r = mid-1;
			}
		}
		for(int i = 1; i <= m; i++)
			out.print(aaa[i]+" ");
	}

    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 = 100000+5;
	static int M = 101;
	DecimalFormat df=new DecimalFormat("0.0000");
	static int inf = (int)1e8;
	static long inf64 = (long) 1e18*2;
	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 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);}
}

CodeForces 514D R2D2 and Droid Army RMQ+二分

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原文地址：http://blog.csdn.net/qq574857122/article/details/44007327