uva 1419 Ugly Windows

题目链接: 点击打开链接

题目大意: 输入几个窗口,判断在最顶上的窗口是哪些

思路: 字符串遍历

分析: 

思路应该没问题.poj上是AC了,但是uva上提交了半天都是wa,上网找了几个代码交上去也是wa...从uva上找了几道之前AC的题,原本可以的代码也都wa了...是不是出了什么问题?

1. 第一步先把不可能在顶端的窗口过滤掉.观察在边界上的字母,如果没被覆盖,周围一定有两个相同字母,找不到说明有窗口overlap了...

2. 还有一种可能是一个小窗口完全叠加在一个大窗口上.这种情况下只要扫描大窗口内部,发现一个不是'.'的字母就可以排除了.

代码:

#include <cstdio>#include <memory.h>using namespace std;const int maxn = 100 + 10;char ui[maxn][maxn];int n, m;bool possible[27], input[27];struct area{int x1, y1, x2, y2;} areas[27];bool inside(int x, int y){return 0 <= x && x < n && 0 <= y && y < m;}bool is_cutted(int x, int y){char c = ui[x][y];int total = 0;if (inside(x-1, y) && ui[x-1][y] == c)++total;if (inside(x+1, y) && ui[x+1][y] == c)++total;if (inside(x, y-1) && ui[x][y-1] == c)++total;if (inside(x, y+1) && ui[x][y+1] == c)++total;return total < 2;}bool is_upper_left(int x, int y){char c = ui[x][y];return inside(x, y+1) && ui[x][y+1] == c && inside(x+1, y) && ui[x+1][y] == c;}bool is_lower_right(int x, int y){char c = ui[x][y];return inside(x, y-1) && ui[x][y-1] == c && inside(x-1, y) && ui[x-1][y] == c;}void init(){for (int i = 0; i < n; ++i)scanf("%s", ui[i]);//for (int j = 0; j < m; ++j) {//scanf("%c", &ui[i][j]);//printf("%c", ui[i][j]);//}memset(possible, 1, sizeof(possible));memset(input, 0, sizeof(input));}void find_cutted(){for (int i = 0; i < n; ++i)for (int j = 0; j < m; ++j) {char c = ui[i][j];if (c!= '.'){input[c - 'A'] = true;if (possible[c - 'A']) {if (is_upper_left(i, j)) {areas[c - 'A'].x1 = i;areas[c - 'A'].y1 = j;}else if (is_lower_right(i, j)) {areas[c - 'A'].x2 = i;areas[c - 'A'].y2 = j;}if (is_cutted(i, j))possible[c - 'A'] = false;}}}}bool scan_inside(int p){char c = p + 'A';int x1 = areas[p].x1, y1 = areas[p].y1, x2 = areas[p].x2, y2 = areas[p].y2;for (int i = x1 + 1; i < x2; ++i)for (int j = y1 + 1; j < y2; ++j)if (ui[i][j] != '.')return false;return true;}void solve(){for (int i = 0; i < 26; ++i)if (input[i] && possible[i] && scan_inside(i))printf("%c", i + 'A');printf("\n");}int main(){while (scanf("%d %d", &n, &m) && n != 0 &&& m != 0){init();find_cutted();solve();}return 0;}