God Save the i-th Queen
Time Limit: 5000ms
Memory Limit: 65536KB
64-bit integer IO format: %lld Java class name: Main
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Tag it!
Did you know that during the ACM-ICPC World Finals a big chessboard is installed every year and is available for the participants to play against each other? In this problem, we will test your basic chess-playing abilities to verify that you would not make a fool of yourself if you advance to the World Finals.
During the yesterday’s Practice Session, you tried to solve the problem of N independent rooks. This time, let’s concentrate on queens. As you probably know, the queens may move not only
horizontally and vertically, but also diagonally.
You are given a chessboard with i−1 queens already placed and your task is to find all squares that may be used to place the i-th queen such that it cannot be captured by any of the others.
Input
The input consists of several tasks. Each task begins with a line containing three integer numbers separated by a space: X, Y , N. X and Y give 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 are N lines, each containing two numbers xk, yk separated by a space. They give the position of the k-th queen, 1 ≤ xk ≤ X, 1 ≤ yk ≤ Y . You may assume that those positions are distinct, i.e., no two queens share the same square.
The last task is followed by a line containing three zeros.
Output
For each task, output one line containing a single integer number: the number of squares which are not occupied and do not lie on the same row, column, or diagonal as any of the existing queens.
Sample Input
8 8 2 4 5 5 5 0 0 0
Sample Output
20 解题思路：刚拿到题目的时候用的暴力，结果数组超内存，又用了set，又超时。后来知道，可以只开4个数组来存覆盖情况。即row，col，pie，na数组来记录行列和撇捺（对角线情况）。可以发现pie数组由x,y相加减1后得到。na数组可以将y转化为相对于右上角的位置为（Y-y+1）。然后枚举地图中各个点，然后判断该点既不在行列，也不在撇捺（对角线）的情况，记录个数即可。
#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,0,sizeof(row));
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; //记录该行被覆盖
col[y]=1; //记录该列被覆盖
pie[x+y-1]=1; //记录右上到左下的对角线被覆盖
na[Y-y+x]=1; //记录左上到右下的对角线被覆盖
}
int num=0;
for(int i=1;i<=X;i++){
for(int j=1;j<=Y;j++){
if((!row[i])&&(!col[j])&&(!pie[i+j-1])&&(!na[Y-j+i])){
//枚举各个点，如果行列撇捺都没覆盖，加1
num++;
}
}
}
printf("%d\n",num);
}
return 0;
}