UVA 784

题目大意：给一个图形，每个图形中用字母分为一块一块，如果一块中有‘*’，即该块内容都填补为‘#’。每个图形下面有下划线作为分割，输出时也要有。

解题思路：类似迷宫问题，将所有‘*’号读入队列，并转化字符。然后bfs。(不看题目坑死人。。。)

ac代码:

#include <iostream>#include <cstring>#include <queue>using namespace std;queue <int>qu;int dx[4] = {-1, 1, 0, 0}, dy[4] = {0, 0, -1, 1};int n, le, wi, vis[1005][1005];char a[1005][1005];void bfs(){while (!qu.empty()){int x, y, len;x = qu.front();qu.pop();y = qu.front();qu.pop();for (int i=0; i<4; i++){int tt, kk;tt = x + dx[i];kk = y + dy[i];len = strlen(a[tt]);if (tt >= 0 && tt < le && kk >= 0 && kk < len && vis[tt][kk] == -1){a[tt][kk] = '#';qu.push(tt);qu.push(kk);vis[tt][kk] = 0;}}}}int main(){scanf("%d", &n);getchar();while (n--){for (int i=0; ;i++){gets(a[i]);if (a[i][0] == '_'){le = i;break;}}memset(vis, -1, sizeof(vis));while (!qu.empty())qu.pop();for (int i=0; i<le; i++){wi = strlen(a[i]);for (int j=0; j<wi; j++){if (isalpha(a[i][j]))vis[i][j] = 0;else if (a[i][j] == '*')a[i][j] = '#', qu.push(i), qu.push(j);}}bfs();for (int i=0; i<le; i++)printf("%s\n", a[i]);printf("%s\n", a[le]);}return 0;}