2018牛客网暑期ACM多校训练营（第二场）I- car ( 思维)
2018牛客网暑期ACM多校训练营（第二场）I- car
链接：https://ac.nowcoder.com/acm/contest/140/I来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld
题目描述
White Cloud has a square of n*n from (1,1) to (n,n).
White Rabbit wants to know the maximum number of cars that can be put into to ensure that there is a way that allows all cars to perform their entire journey without damage.
(update: all cars should start at the edge of the square and go towards another side, cars which start at the corner can choose either of the two directions)
For example, in a 5*5 square
legal
illegal（These two cars will collide at (4,4)）
illegal (One car will go into a damaged grid)
输入描述:
The first line of input contains two integers n and m(n <= 100000,m <= 100000)For the next m lines,each line contains two integers x,y(1 <= x,y <= n), denoting the grid which is damaged by White Cloud.
输出描述:
Print a number,denoting the maximum number of cars White Rabbit can put into.
示例1
输入
[复制](javascript:void(0)😉
2 0
输出
[复制](javascript:void(0)😉
4
备注:
思路：
最优的放车方案为如上图所示。
所以我们只需要开2个数组，分别 $c[i] $ 代表第i行是否有坑，$d[i] $ 代表第i列是否有坑，
然后再判断下，如果N是奇数，中间的十字路只能放一个车，细节注意下即可。
遇到这种问题主要是应该冷静分析最优的方案，题目中给的参数对最优方案有什么影响？
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
map<int, bool> vis;
int n, m;
int getid(int x, int y)
{
return n * (x - 1) + y;
}
pii a[maxn];
bool hang[maxn];
bool lie[maxn];
int main()
{
//freopen("D:\\code\\text\\input.txt","r",stdin);
//freopen("D:\\code\\text\\output.txt","w",stdout);
du2(n, m);
int x, y;
repd(i, 1, m) {
du2(x, y);
hang[x] = 1;
lie[y] = 1;
}
if (n == 1 && m == 0) {
return 0 * puts("1");
} else if (n == 1) {
return 0 * puts("0");
}
int ans = 0;
for (int i = 1; i <= n; ++i) {
if (!lie[i]) {
ans++;
}
}
for (int i = 1; i <= n; ++i) {
if (n & 1) {
if ((i + i - 1) == n) {
continue;
}
}
if (!hang[i]) {
ans++;
}
}
printf("%d\n", ans );
return 0;
}
inline void getInt(int *p)
{
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
} else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}
