B. Chips
http://codeforces.com/problemset/problem/333/B
time limit per test
memory limit per test
input
output
n × n cells, where m various cells are banned. Before the game, he has to put a few chips on some border (but not corner) board cells. Then for n - 1
- At least one of the chips at least once fell to the banned cell.
- At least once two chips were on the same cell.
- At least once two chips swapped in a minute (for example, if you stand two chips on two opposite border cells of a row with even length, this situation happens in the middle of the row).
0
Input
n and m (2 ≤ n ≤ 1000, 0 ≤ m ≤ 105) — the size of the field and the number of banned cells. Next m lines each contain two space-separated integers. Specifically, the i-th of these lines contains numbers xi and yi (1 ≤ xi, yi ≤ n) — the coordinates of the i-th banned cell. All given cells are distinct.
n, and the columns — from left to right from 1 to n.
Output
Print a single integer — the maximum points Gerald can earn in this game.
Sample test(s)
input
3 1 2 2
output
0
input
3 0
output
1
input
4 3 3 1 3 2 3 3
output
1
Note
In the first test the answer equals zero as we can't put chips into the corner cells.
In the second sample we can place one chip into either cell (1, 2), or cell (3, 2), or cell (2, 1), or cell (2, 3). We cannot place two chips.
In the third sample we can only place one chip into either cell (2, 1), or cell (2, 4).
简单贪心，考虑相邻的两条边即可，详见代码。
完整代码：（未优化）
/*92ms,0KB*/
#include<cstdio>
bool isban[2][1000];
int main()
{
int n, m, x, y, count = 0;
scanf("%d%d", &n, &m);
while (m--)
{
scanf("%d%d", &x, &y);
isban[0][x] = true;
isban[1][y] = true; ///只统计相邻的两条边就行
}
for (int i = 2; i < n; ++i)
{
if (!isban[0][i])
count++;
if (!isban[1][i])
count++;
}
///在这种情况下两个chips(芯片/小木片)重合，上面多算了一次count
if (n & 1 && !(isban[0][(n + 1) >> 1] || isban[1][(n + 1) >> 1]))
count--;
printf("%d", count);
return 0;
}优化版：
/*62ms,0KB*/
#include<cstdio>
bool isban[2][1000];
int main()
{
int n, m, x, y;
scanf("%d%d", &n, &m);
int count = (n << 1) - 4;
isban[0][1]=true;
isban[0][n]=true;
isban[1][1]=true;
isban[1][n]=true;
while (m--)
{
///只统计相邻的两条边就行
scanf("%d%d", &x, &y);
if (!isban[0][x])
{
--count;
isban[0][x] = true;
}
if (!isban[1][y])
{
--count;
isban[1][y] = true;
}
}
///在这种情况下两个chips(芯片/小木片)重合，上面多算了一次count
if (n & 1 && !(isban[0][(n + 1) >> 1] || isban[1][(n + 1) >> 1]))
--count;
printf("%d", count);
return 0;
}
