新博客地址: vonsdite.cn
题目网址: Educational Codeforces Round 26 C. Two Seals
题意分析:
题意:
就是求取出两个矩形放在已知的矩形区域内, 使得找出的这两个矩形占的面积最大粗暴一点… 把八种情况都写了…
代码:
#include <bits/stdc++.h>
using namespace std;
const int SIZE = 105;
pair<int, int> point[SIZE];
int main(int argc, char const *argv[])
{
int n, a, b;
int x, y;
while (~scanf("%d %d %d", &n, &a, &b))
{
for (int i = 0; i < n; ++i)
{
scanf("%d %d", &x, &y);
point[i] = make_pair(x, y);
}
int MAX = 0;
for (int i = 0; i < n; ++i)
{
for (int j = i + 1; j < n; ++j)
{
if (point[i].first + point[j].first <= a \
&& point[i].second <= b\
&& point[j].second <= b)
{
MAX = max(MAX, point[i].first*point[i].second \
+ point[j].first*point[j].second);
}
if(point[i].first + point[j].second <= a \
&& point[i].second <= b \
&& point[j].first <= b)
{
MAX = max(MAX, point[i].first*point[i].second \
+ point[j].first*point[j].second);
}
if (point[i].first + point[j].first <= b \
&& point[i].second <= a\
&& point[j].second <= a)
{
MAX = max(MAX, point[i].first*point[i].second \
+ point[j].first*point[j].second);
}
if(point[i].first + point[j].second <= b \
&& point[i].second <= a \
&& point[j].first <= a)
{
MAX = max(MAX, point[i].first*point[i].second \
+ point[j].first*point[j].second);
}
// 5
if (point[i].second + point[j].first <= a \
&& point[i].first <= b\
&& point[j].second <= b)
{
MAX = max(MAX, point[i].first*point[i].second \
+ point[j].first*point[j].second);
}
if(point[i].second + point[j].second <= a \
&& point[i].first <= b \
&& point[j].first <= b)
{
MAX = max(MAX, point[i].first*point[i].second \
+ point[j].first*point[j].second);
}
if (point[i].second + point[j].first <= b \
&& point[i].first <= a\
&& point[j].second <= a)
{
MAX = max(MAX, point[i].first*point[i].second \
+ point[j].first*point[j].second);
}
if(point[i].second + point[j].second <= b \
&& point[i].first <= a \
&& point[j].first <= a)
{
MAX = max(MAX, point[i].first*point[i].second \
+ point[j].first*point[j].second);
}
}
}
printf("%d\n", MAX);
}
return 0;
}
