UVa 12714 Two Points Revisited (水题，计算几何)

题意：给定一条线段，让你求一条线段与已知线段垂直，并且所有线段的坐标的点的坐标都不大于给定的坐标的最大值且不能为负数。

析：没啥好说的，随便找一条就好。

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>
//#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 = 2500 + 5;
const LL mod = 10000000000007;
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); }
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(){
    int x1, y1, x2, y2, T;
    cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        cin >> x1 >> y1 >> x2 >> y2;
        m = 0;
        m = Max(m, x1);  m = Max(m, y1);
        m = Max(m, x2);  m = Max(m, y2);
        int x = x2 - x1;
        int y = y2 - y1;
        int xx = -y, yy = x;
        int xxx = xx + x1, yyy = yy + y1;
        if(xxx < 0){
            x1 -= xxx;
            xxx = 0;
        }
        if(yyy < 0){
            y1 -= yyy;
            yyy = 0;
        }
        if(xxx > m){
            x1 += (m - xxx);
            xxx = m;
        }
        if(yyy > m){
            y1 += (m - yyy);
            yyy = m;
        }
        printf("Case %d: %d %d %d %d\n", kase, x1, y1, xxx, yyy);

    }
    return 0;
}