Hihocoder 太阁最新面经算法竞赛18

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;
}