Educational Codeforces Round 26 C
题意:
思路:模拟(枚举第一个的2种摆法,暴力第二个看是否可以放下,前提是第一个可以放下)
AC代码:
#include "iostream" #include "iomanip" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #pragma comment(linker, "/STACK:102400000,102400000") #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a,x) memset(a,x,sizeof(a)) #define step(x) fixed<< setprecision(x)<< #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define ll long long #define endl ("\n") #define ft first #define sd second #define lrt (rt<<1) #define rrt (rt<<1|1) using namespace std; const long long INF = 1e18+1LL; const int inf = 1e9+1e8; const int N=1e5+100; const ll mod=1e9+7; int n,a,b,ans,x[105], y[105]; int main(){ ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); cin>>n>>a>>b; if(a>b) swap(a,b); for(int i=1; i<=n; ++i){ cin>>x[i]>>y[i]; if(x[i]>y[i]) swap(x[i],y[i]); } int h1,l1,h2,l2; for(int i=1; i<=n; ++i){ for(int j=i+1; j<=n; ++j){ if(x[i]<=a && y[i]<=b){ h1=a-x[i], l1=b; if(h1>l1) swap(h1,l1); h2=a, l2=b-y[i]; if(h2>l2) swap(h2,l2); if((x[j]<=h1 && y[j]<=l1) || (x[j]<=h2 && y[j]<=l2)){ ans=max(ans,x[i]*y[i]+x[j]*y[j]); } } if(x[i]<=b && y[i]<=a){ h1=a-y[i], l1=b; if(h1>l1) swap(h1,l1); h2=a, l2=b-x[i]; if(h2>l2) swap(h2,l2); if((x[j]<=h1 && y[j]<=l1) || (x[j]<=h2 && y[j]<=l2)){ ans=max(ans,x[i]*y[i]+x[j]*y[j]); } } } } cout<<ans<<endl; return 0; }