Gym 100531G Grave(水题)

Posted 2020-08-15 dwtfukgv

题意：给定一个大矩形，再给定在一个小矩形，然后给定一个新矩形的长和高，问你能不能把这个新矩形放到大矩形里，并且不与小矩形相交。

析：直接判定小矩形的上下左右四个方向，能不能即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50000 + 5;
const LL mod = 1e3 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
    freopen("grave.in", "r", stdin);
    freopen("grave.out", "w", stdout);
    int x1, y1, x2, y2, x3, y3, w, h, x4, y4;
    while(cin >> x1 >> y1 >> x2 >> y2){
        cin >> x3 >> y3 >> x4 >> y4;
        cin >> w >> h;
        bool ok = false;
        if(y2 - y4 >= h && x2 - x1 >= w)  ok = true;
        else if(y3 - y1 >= h && x2 - x1 >= w)  ok = true;
        else if(x3 - x1 >= w && y2 - y1 >= h)  ok = true;
        else if(x2 - x4 >= w && y2 - y1 >= h)  ok = true;
        printf("%s\n", ok ? "Yes" : "No");
    }
    return 0;
}