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
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
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; }