cf1555B. Two Tables

题意：

一个大矩阵空间内放置一个矩阵a，现在要再往这个空间内放一个矩阵b，a移动距离len才能放下b，问len最小是多少

题解：

不难发现左右或上下移动是最佳的，斜着移动是最不好的。此时我们就需要判断左右移动是否有足够的空间，并且移动后是否可以安放b
b的安防一定是靠着墙角的

代码：

// Problem: B. Two Tables
// Contest: Codeforces - Educational Codeforces Round 112 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1555/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Data:2021-08-16 23:47:26
// By Jozky#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read()
{
}
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{x= 0;char c= getchar();bool flag= 0;while (c < '0' || c > '9')flag|= (c == '-'), c= getchar();while (c >= '0' && c <= '9')x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();if (flag)x= -x;read(Ar...);
}template <typename T> inline void write(T x)
{if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
int main()
{//rd_test();int t;read(t);while (t--) {int a, b, x1, x2, y1, y2, a1, b1, a2, b2;read(a);read(b);cin >> x1 >> y1 >> x2 >> y2 >> a2 >> b2;a1= x2 - x1;//矩阵a的长b1= y2 - y1;//矩阵a的宽if (a1 + a2 > a && b1 + b2 > b) {//如果矩阵a和矩阵b长宽加起来会超出空间printf("-1\n"); //超界continue;}int ans= 0x3f3f3f3f;if (a1 + a2 <= a) { //横向移动不会出界if (x1 >= a2 || x2 + a2 <= a)ans= 0; //不用移动elseans= min(ans, min(a2 - x1, x2 + a2 - a)); //左右移动取最小}if (b1 + b2 <= b) {if (y1 >= b2 || y2 + b2 <= b)ans= 0;elseans= min(ans, min(b2 - y1, y2 + b2 - b));}printf("%d.000000000\n", ans);}return 0;//Time_test();
}