USACO 1.2.2 Transformations(transform)

最新推荐文章于 2023-12-29 17:52:45 发布

原创最新推荐文章于 2023-12-29 17:52:45 发布 · 821 阅读

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本文探讨了一个关于矩阵旋转和翻转的问题，在特定条件下如何通过多次操作使两个矩阵匹配。涉及不同操作步骤的实现及调试过程。

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1、为什么MAX等于10以下数据在我的机器上结果正确，而提交通不过？换成11就可以。有谁知道了麻烦留言告诉一声。：）

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/* ID: gengjia1 LANG: C TASK: transform */ #include <stdio.h> #include <stdlib.h> #include <string.h> //#define NDEBUG #include <assert.h> #define MAX 11 //10通不过？？ //顺时针旋转90度公式a[i][j]->a[j][N-i-1] void rotate90(char a[][MAX],int N) { char tmp[MAX][MAX]; int i, j; for(i = 0; i < N; i++) { for(j = 0; j < N; j++) { tmp[j][N-i-1] = a[i][j]; } } for(i = 0; i < N; i++) { for(j = 0; j < N; j++) { a[i][j] = tmp[i][j]; } } } //水平翻转公式a[i][j]->a[i][N-j-1] void reflect(char a[][MAX], int N) { char tmp; int i, j; for(i = 0; i < N; i++) { for(j = 0; j < N/2; j++) { tmp = a[i][N-j-1]; a[i][N-j-1] = a[i][j]; a[i][j] = tmp; } } } int compare(char b[][MAX], char a[][MAX], int N) { int i, j; for(i = 0; i < N; i++) { for(j = 0; j < N; j++) { if(b[i][j] != a[i][j]) { return 1; //不相等返回1 } } } return 0; } int main(void) { FILE *fin = fopen ("transform.in", "r"); FILE *fout = fopen ("transform.out", "w"); int N, count = 0; char before[MAX][MAX], after[MAX][MAX]; char tb[MAX][MAX]; int i, j; fscanf (fin, "%d", &N); assert(N >= 1); assert(N <= MAX); for(i = 0; i < N; i++) { fscanf (fin, "%s", before[i]); } for(i = 0; i < N; i++) { fscanf (fin, "%s", after[i]); } for(i = 0; i < N; i++) { for(j = 0; j < N; j++) { tb[i][j] = before[i][j]; } } rotate90(before, N); if(compare(before, after, N) == 0) { count = 1; } else if(rotate90(before, N), compare(before, after, N) == 0) { count = 2; } else if(rotate90(before, N), compare(before, after, N) == 0) { count = 3; } else { for(i = 0; i < N; i++) { for(j = 0; j < N; j++) { before[i][j] = tb[i][j]; } } reflect(before, N); if(compare(before, after, N) == 0) { count = 4; } else if(rotate90(before, N), compare(before, after, N) == 0) { count = 5; } else if(rotate90(before, N), compare(before, after, N) == 0) { count = 5; } else if(rotate90(before, N), compare(before, after, N) == 0) { count = 5; } } if(count == 0) { for(i = 0; i < N; i++) { for(j = 0; j < N; j++) { before[i][j] = tb[i][j]; } } if(compare(before, after, N) == 0) { count = 6; } else { count = 7; } } fprintf (fout, "%d/n", count); fclose(fin); fclose(fout); exit(0); }