标签：isp turn ons chess his name view spec vector
Vasya has the square chessboard of size n?×?n and m rooks. Initially the chessboard is empty. Vasya will consequently put the rooks on the board one after another.
The cell of the field is under rook‘s attack, if there is at least one rook located in the same row or in the same column with this cell. If there is a rook located in the cell, this cell is also under attack.
You are given the positions of the board where Vasya will put rooks. For each rook you have to determine the number of cells which are not under attack after Vasya puts it on the board.
The first line of the input contains two integers n and m (1?≤?n?≤?100?000, 1?≤?m?≤?min(100?000,?n2)) — the size of the board and the number of rooks.
Each of the next m lines contains integers xi and yi (1?≤?xi,?yi?≤?n) — the number of the row and the number of the column where Vasya will put the i-th rook. Vasya puts rooks on the board in the order they appear in the input. It is guaranteed that any cell will contain no more than one rook.
Print m integer, the i-th of them should be equal to the number of cells that are not under attack after first irooks are put.
#include <cstdio> #include <cctype> #include <stdlib.h> #include <iostream> #include <cmath> #include <iomanip> #include <cstring> #include <algorithm> #include <string> #include <vector> #include <map> using namespace std; typedef long long LL; int row[100005] = {0}; int col[100005] = {0}; LL n,m; LL xtotal = 0,ytotal = 0; int main() { // freopen("test.in","r",stdin); cin >> n >> m; LL sum = n * n; for (int i=1;i<=m;i++){ int x,y; cin >> x >> y; if (!row[x]){ sum -= (n - xtotal); row[x] = 1; ytotal ++; } if (!col[y]){ sum -= (n - ytotal); col[y] = 1; xtotal ++; } cout << sum << " "; } return 0; }
Codeforces 701B. Cells Not Under Attack
标签：isp turn ons chess his name view spec vector
原文地址：http://www.cnblogs.com/ToTOrz/p/6875675.html
