Bnu4299--god Save the i-th Queen —————— "Queen attack, find the corresponding compression space"

God Save the i-th queentime limit:5000msmemory limit:65536kb64-bit integer IO format:%lld Java class name: Main Prev Submit Status Statistics discuss nextdid you know this during the ACM-ICPC World Finals a big chessboard is installed Every year and was available for the participants to play against each other? In this problem, we'll test your basic chess-playing abilities to verify so you would don't make a fool of yourself if Y OU advance to the world finals.during the yesterday's practice Session, you tried to solve the problem of N Independent ro OKs. This time, let's concentrate on Queens. As you probably know, the Queens could move not onlyhorizontally and vertically, but also diagonally. You were given a chessboard with i−1 queens already placed and your task was to find all squares the May was used to place t He i-th queen such that it cannot is captured by any of the others. Inputthe input consists of several tasks. Each task begins with a line containing three integer numbers separated by a space: X, Y, N. Xand YGive the chessboard size, 1 ≤ X, Y ≤20 000. N= I −1 is the number of Queens already placed, 0 ≤ N ≤ X · Y. After the first line, there is NLines, each containing, numbers XK, YKseparated by a space. They give the position of the k-th Queen, 1 ≤ XK ≤ X, 1 ≤ YK ≤ Y. Assume that those positions is distinct, i.e., No. Queens share the same square. The last task was followed by a line containing three zeros. Outputfor each task, output one line containing a single integer number:the number of squares which is not occupied and Do not lie on the same row, column, or diagonal as any of the existing queens. Sample Input

8 8 24 55 50 0 0

Sample Output

20


Problem-solving ideas: Just get the problem with the violence, the result of the array of super memory, and set, and time out. Later, you can only open 4 arrays to save coverage. That is, the Row,col,pie,na array to record rows and apostrophes (diagonal cases). You can see that the pie array is added by X, Y and minus 1. The NA array can convert Y to the position relative to the upper-right corner (y-y+1). Then enumerate the points in the map, and then determine whether the point is not in the row, nor the apostrophe (diagonal), the number of records can be.

#include <bits/stdc++.h>using namespace Std;const int Maxn=21000;bool row[maxn],col[maxn],pie[maxn*2],na[maxn*    2];void init () {memset (row) (row,0,sizeof);    memset (col,0,sizeof (col));    memset (pie,0,sizeof (PIE)); memset (na,0,sizeof (NA));}    int main () {int x,y,n;        while (scanf ("%d%d%d", &x,&y,&n)!=eof&& (x+y+n)) {init ();            for (int i=0;i<n;i++) {int x, y;            scanf ("%d%d", &x,&y);       Row[x]=1;       Record that the line is overwritten col[y]=1;   Record that the column is overwritten pie[x+y-1]=1;    The top right-to-left diagonal of the record is covered na[y-y+x]=1;        Record left top to bottom right diagonal is overwritten} int num=0; for (int i=1;i<=x;i++) {for (int j=1;j<=y;j++) {if ((!row[i]) && (!col[j]) && (!p                Ie[i+j-1]) && (!na[y-j+i]) {//Enumerate each point, if the row and column is not covered, add 1 num++;    }}} printf ("%d\n", num); } return 0;}

　　

Bnu4299--god Save the i-th Queen —————— "Queen attack, find the corresponding relationship compression space"