1.Splitting the Field

http://usaco.org/index.php?page=viewproblem2&cpid=645

給二維坐標系中的n個點，求ans=用一個矩形覆蓋所有點所用矩形面積-用兩個矩形覆蓋所有點所用兩個矩形的最小面積和，而且兩個矩形不能重合(邊重合也不行)

枚舉兩個矩形的分割線，也就是把所有點分成兩個部分，枚舉分割點；先預處理每個點之前和之後的最大，最低高度

#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cstring>
#define
 
 N 50015
#define down(i,r,l) for(int i=r;i>=l;i--)
#define rep(i,l,r) for(int i=l;i<=r;i++)
using namespace std;
typedef long long ll;
int l[N],r[N],l1[N],r1[N],n;
ll zs,ans;
struct node{int x,y;}s[N];
bool cmp(node x,node y) {return x.x==y.x?x.y<y.y:x.x<y.x;}
void getans() { 
     ll now;
     sort(s
 
+1,s+1+n,cmp);
     memset(l,60,(n+10)<<2); memset(l1,60,(n+10)<<2); memset(r,0,(n+10)<<2); memset(r1,0,(n+10)<<2);
     rep(i,1,n) l[i]=min(l[i-1],s[i].y),r[i]=max(r[i-1],s[i].y);
     down(i,n,1) l1[i]=min(l1[i+1],s[i].y),r1[i]=max(r1[i+1],s[i].y);
     rep(i,2,n) {
          now=(ll)(s[i-1
 
].x-s[1].x) * (ll)(r[i-1]-l[i-1]) + (ll)(s[n].x-s[i].x) * (ll)(r1[i]-l1[i]); 
          ans=min(ans,now);
     }
}
int main () 
{
     int miny,minx,maxx,maxy;
     miny=minx=1000000050,maxy=maxx=0;
     scanf("%d",&n); rep(i,1,n) 
     {
         scanf("%d%d",&s[i].x,&s[i].y),miny=min(miny,s[i].y),maxy=max(maxy,s[i].y);;
         maxx=max(maxx,s[i].x); minx=min(minx,s[i].x);
     }
     ans=(ll)(maxy-miny)*(ll)(maxx-minx),zs=ans;
     getans();
     rep(i,1,n) swap(s[i].x,s[i].y);
     getans(); 
     printf("%lld\n",zs-ans);
}

2.Closing the Farm

http://usaco.org/index.php?page=viewproblem2&cpid=646

離線+並查集

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#define N 200020
#include<vector>
using namespace std;
int n,m,a[N],f[N];
bool vis[N],ok[N];
vector<int>G[N];
int find(int x){return x==f[x]?x:f[x]=find(f[x]);}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1,a,b;i<=m;i++)
    {
        scanf("%d%d",&a,&b);
        G[a].push_back(b);
        G[b].push_back(a);
    }
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]),f[i]=i;
    int cnt=0;
    for(int i=n;i;i--)
    {
        int u=a[i];
        cnt++;vis[u]=1;
        for(int j=0;j<G[u].size();j++)
        {
            int v=G[u][j];
            if(!vis[v])continue;
            int fa=find(u),fb=find(v);
            if(fa!=fb)
            {
                if(fa>fb)swap(fa,fb);
                f[fb]=fa;
                cnt--;
            }
        }
        if(cnt==1)ok[i]=1;
    }
    for(int i=1;i<=n;i++)
        if(ok[i])printf("YES\n");
        else printf("NO\n");
}

3.248

http://usaco.org/index.php?page=viewproblem2&cpid=647#

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=3e5+10;
int n,ans,a,f[60][maxn];
int main()
{
      scanf("%d",&n);
      for(int i=1;i<=n;i++)
    {
        scanf("%d",&a);
        f[a][i]=i+1;
    }
      for(int i=2;i<=58;i++)
           for(int j=1;j<=n;j++)
           {
            if(!f[i][j])f[i][j]=f[i-1][f[i-1][j]];
             if(f[i][j])ans=max(i,ans);
           }
      printf("%d",ans);
      return 0;
} 

USACO 2016 US Open Contest, Gold解題報告