uva 159 wrod crosses

不用复杂的算法，但是有比较复杂的处理。

 

#include #include #include #include using namespace std; #undef _DEBUG #ifdef _DEBUG #define PRINT_VAR() {/ printf("%3d: p = %x, q = %x/n", __LINE__, p, q);/ } #endif char ans[30][50]; char a[100], b[100], c[100], d[100]; int main() { int cs = 1; while(scanf("%s%s%s%s", a, b, c, d) == 4) { memset(ans, 0, sizeof(ans)); strcpy(ans[10], a); int offset = strlen(a) + 3; strcpy(ans[10] + offset, c); char *p = a; while(*p) { char *q = b - 1; while(*++q && *q != *p) {} if(*q) { int i = 1; char *t = q; while(--t >= b) ans[10 - i++][p - a] = *t; t = q; i = 1; while(*++t) ans[10 + i++][p - a] = *t; break; } #ifdef _DEBUG PRINT_VAR(); #endif ++p; } bool isSolvable = ((*p) ? true : false); p = c; while(*p && isSolvable) { char *q = d - 1; while(*++q && *q != *p) {} if(*q) { int i = 1; char *t = q; while(--t >= d) ans[10 - i++][p - c + offset] = *t; t = q; i = 1; while(*++t) ans[10 + i++][p - c + offset] = *t; break; } #ifdef _DEBUG PRINT_VAR(); #endif ++p; } isSolvable = isSolvable && (*p); if(cs++ != 1) printf("/n"); if(! isSolvable) { printf("Unable to make two crosses/n"); continue; } int r = 0; while(r < 30) { bool lineIsEmpty = true; int c = 0; while(c < 50) { int spaces = 0; while(c < 50 && !ans[r][c]) { ++c; ++spaces; } if(c >= 50) { if(! lineIsEmpty) printf("/n"); ++c; continue; } lineIsEmpty = false; while(spaces-- > 0) #ifdef _DEBUG printf("."); #else printf(" "); #endif printf("%c", ans[r][c]); ++c; } ++r; } } return 0; }