Hihocoder 太阁最新面经算法竞赛18

Posted 2020-08-25 zhang--yd

Hihocoder 太阁最新面经算法竞赛18

source: https://hihocoder.com/contest/hihointerview27/problems

 

题目1 : Big Plus

描述

Given an NxN 01 matrix, find the biggest plus (+) consisting of 1s in the matrix.

size 1 plus   size 2 plus   size 3 plus  size 4 plus
     1                1              1                1
    111               1              1                1
     1              11111            1                1
                      1           1111111             1
                      1              1            111111111
                                     1                1
                                     1                1
                                                      1
                                                      1

输入

The first line contains an integer N. (1 <= N <= 500)  

Then follow an NxN 01 matrix.

输出

The size of the biggest plus in the matrix.

样例输入5  

00100  
00100  
11111  
00110  
10101

样例输出  2

 

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> 
using namespace std;
const int MAXN = 505;

int n, ans; 
char mp[MAXN][MAXN];
int U[MAXN][MAXN], D[MAXN][MAXN], LEF[MAXN][MAXN], RIG[MAXN][MAXN]; 

int main(){
	int tmp; 
	while(scanf("%d", &n) != EOF){
		for(int i=1; i<=n; ++i){
			getchar(); 
			for(int j=1; j<=n; ++j){
				scanf("%c", &mp[i][j]);
			}
		}
		memset(U, 0, sizeof(U)); memset(D, 0, sizeof(D)); 
		memset(LEF, 0, sizeof(LEF)); memset(RIG, 0, sizeof(RIG)); 
		for(int i=1; i<=n; ++i){
			for(int j=1; j<=n; ++j){
				if(mp[i][j] == ‘1‘){
					U[i][j] = U[i-1][j] + 1; 
					LEF[i][j] = LEF[i][j-1] + 1; 
				}
			}
		}
		for(int i=n; i>=1; --i){
			for(int j=n; j>=1; --j){
				if(mp[i][j] == ‘1‘){
					D[i][j] = D[i+1][j] + 1; 
					RIG[i][j] = RIG[i][j+1] + 1; 
				}
			}
		}
		ans = 0; 
		for(int i=1; i<=n; ++i){
			for(int j=1; j<=n; ++j){
				if(mp[i][j] == ‘1‘){
					tmp = min(min(U[i][j], RIG[i][j]), min(D[i][j], LEF[i][j])); 
					if(tmp-1 > ans){
						ans = tmp - 1; 
					}
				}
			}
		}
		printf("%d\n", ans );
	}
	return 0; 
}

　　

 

题目2 : Interval Coverage

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

You are given N intervals [S1, T1], [S2, T2], [S3, T3], ... [SN, TN] and a range [X, Y]. Select minimum number of intervals to cover range [X, Y].

输入

The first line contains 3 integers N, X and Y. (1 <= N <= 100000, 1 <= X < Y <= 1000000)

The following N lines each contain 2 integers Si, Ti denoting an interval. (1 <= Si < Ti <= 1000000)

输出

Output the minimum number of intevals to cover range [X, Y] or -1 if it is impossible.

样例输入

5 1 5
1 2    
1 3  
2 4  
3 5  
4 5 

样例输出

2

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> 
using namespace std;
const int MAXN = 100000 + 5;
struct Interval{
	int s, t; 
}inte[MAXN]; 
int n, x, y;

int cmp(const void *a, const void *b){
	Interval *aa = (Interval *)a; 
	Interval *bb = (Interval *)b; 
	if(aa->s == bb->s){
		return aa->t - bb->t; 
	}
	return aa->s - bb->s; 
}
int main(){
	freopen("in.txt", "r", stdin); 

	int ans, far, start; 
	while(scanf("%d %d %d", &n, &x, &y) != EOF){
		for(int i=0; i<n; ++i){
			scanf("%d %d", &inte[i].s, &inte[i].t); 
		}
		qsort(inte, n, sizeof(inte[0]), cmp); 
		if(inte[0].s > x){
			ans = -1; 
		}else{
			far = max(x, inte[0].t); 
			start = x; 
			ans = 1; 
			for(int i=0; i<n; ++i){
				if(far >= y){
					break; 
				}
				if(inte[i].s <= start){
					far = max(far, inte[i].t); 
				}else if(inte[i].s > far){
					break; 
				}else{
					start = far; 
					far = max(far, inte[i].t); 
					++ans; 
				}
			}
			if(far < y){
				ans = -1; 
			}
		}
		printf("%d\n", ans );
	}
	return 0; 
}

　　

 

题目3 : Split Array

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

You are given an sorted integer array A and an integer K. Can you split A into several sub-arrays that each sub-array has exactly K continuous increasing integers.

For example you can split {1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6}  into {1, 2, 3}, {1, 2, 3}, {3, 4, 5}, {4, 5, 6}.  

输入

The first line contains an integer T denoting the number of test cases. (1 <= T <= 5)

Each test case takes 2 lines. The first line contains an integer N denoting the size of array A and an integer K. (1 <= N <= 50000, 1 <= K <= N)

The second line contains N integers denoting array A. (1 <= Ai <= 100000)

输出

For each test case output YES or NO in a separate line.

样例输入

2  
12 3 
1 1 2 2 3 3 3 4 4 5 5 6  
12 4  
1 1 2 2 3 3 3 4 4 5 5 6

样例输出

YES  
NO

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> 
using namespace std;
const int MAXN = 50000 + 5;
const int MAXV = 100000 + 5; 

int n, k, num[MAXV]; 
int main(){
	freopen("in.txt", "r", stdin); 

	int test_case, val, maxval, tmp, flag;   
	scanf("%d", &test_case); 

	while(test_case--){
		scanf("%d %d", &n, &k); 
		maxval = 0; 
		memset(num, 0, sizeof(num)); 
		for(int i=0; i<n; ++i){
			scanf("%d", &val); 
			num[val]++; 
			maxval = max(maxval, val); 
		}
		flag = 1; 
		for(int i=1; i<=maxval; ++i){
			if(num[i] > 0){
				tmp = num[i]; 
				for(int j=0; j<k; ++j){
					num[i+j] -= tmp; 
				}
			}else if(num[i] < 0){
				flag = 0;
				break; 
			}
		}
		if(flag){
			printf("YES\n");
		}else{
			printf("NO\n");
		}
	}
	return 0; 
}