Oh, there is a flower and can't open it, and you will be unhappy. . . F question (pku1899) actually found a three-character BUG for 4 hours, or three people watch together. . . And C questions (PKU1886) actually wrote it, and I was AC. . . It's too uncomfortable. If you can't see it next time, you should be separated, and you can't find 5 hours, so you die very miserable. . . Finally, Lijunwei and I have a question, Flybird just later looks F, did not make it. . . Unfortunately. . I monsted me to put bug, and then set them, no one was discovered. . Two questions. . . Let's refuel the next time, we still have a chance, Well 16, strive :)
Attachment (PKU1899) 2004.10.10 The correct answer and 3 characters of 8 hours of 8 hours have not found:
Discover your program problem
#include
#include
Using namespace std;
INT I, J;
Bool Findit (int UP1, INT DOWN1, INT LEFT1, INT RIGHT1,
INT UP2, INT DOWN2, INT LEFT2, INT RIGHT2
{
IF (Down2> = DOWN1 && DOWN2 <= UP1 && Left2> = Left1 && Left2 <= Right1) {
Return True;
}
IF (Down2> = DOWN1 && DOWN2 <= UP1 && Right2> = LEFT1 && Right2 <= Right1) {
Return True;
}
IF (Up2> = DOWN1 && UP2 <= UP1 && Left2> = Left1 && Left2 <= Right1) {
Return True;
}
IF (UP2> = DOWN1 && UP2 <= Up1 && Right2> = Left1 && Right2 <= Right1) {
Return True;
}
Return False;
}
#include
Int main (void)
{
// ifstream cin ("test.txt"); int x, y, n; // 1 ~ 1000xy, 0 ~ 100n
INT R [105] [3];
INT O [105] [5]; // The 0th representation of each line is effective
While (CIN >> X >> Y) {
// input
CIN >> N;
IF (n == 0) {
COUT << x * y << "/ n";
CONTINUE;
}
For (i = 1; i <= n; i ) {
CIN >> R [I] [0] >> R [I] [1] >> R [I] [2];
}
//
For (i = 1; i <= n; i ) {
o [i] [0] = 0;
}
// Cal
INT UP1, DOWN1, LEFT1, RIGHT1
For (INT i = 1; i <= n; i ) {
Up1 = r [i] [1] r [i] [2];
DOWN1 = R [I] [1] - R [i] [2];
LEFT1 = R [I] [0] -r [i] [2];
Right1 = r [i] [0] r [i] [2];
// Take a circle to see if it is intersecting, and the intersection will kill the inside.
For (j = 1; j IF (o [j] [0]! = 0) {// effective, take out judgment / / View two rectangles intersect IF (Findit (O [J], O [J] [2], O [J] [3], O [J] [4], UP1, DOWN1, LEFT1, RIGHT1) || Findit (Up1, Down1, LEFT1, RIGHT1, O [J] [1], O [J] [2], O [J] [3], o [j]])) { //merge IF (O [J] [1]> UP1) UP1 = O [J] [1]; IF (O [J] [2] Down1 = O [J] [2]; IF (O [J] [3] Left1 = O [J] [3]; IF (O [J] [4]> Right1) Right1 = O [J] [4]; O [J] [0] = 0; // ******************************************************** ******************************* J = 0; // BUG, after the new rectangle is expanded, you should search from the beginning, it is possible to extend the front. . // ******************************************************** ******************************* } } } o [i] [0] = 1; o [i] [1] = UP1; o [i] [2] = DOWN1; o [i] [3] = left1; } INT all = x * y; For (i = 1; i <= n; i ) { IF (o [i] [0]! = 0) { All = all- (o [i] [1] -O [i] [2]) * (o [i] [4] -o [i] [3]); } } IF (all <0) ALL = 0; Cout << all << endl; } Return 0; } Bamboo
