US Open 2016 Contest

比较弱，只写了金组和银组，铂金组的第一题。

 

【262144】

http://www.usaco.org/index.php?page=viewproblem2&cpid=648

给一个序列，相邻两个数组相同的可以合并变成数值大一的数，问最大能得到的数。n<262144,val<=40

是金组某道题的升级版。

因为val<40，实际上最大可以取到的不超过60.所以从1-60，看看从i开始是否能合并到当前数值。可以的话记录合并到哪一位。

初始化f[i][val[i]]=i+1,状态转移f[i][j]=f[f[i][j-1][j-1]。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<algorithm>
#include<stack>
#include<cmath>
#include<vector>
#include<map>
#define maxn 300000
using namespace std;

int f[62][maxn],num[maxn],ans,n;

void input() {freopen("262144.in","r",stdin);freopen("262144.out","w",stdout);}
void output() {fclose(stdin);fclose(stdout);}
void work()
{
    int i,k;
    memset(f,0,sizeof(f));
    for (k=0;k<=60;k++) {
        for (i=1;i<=n;i++) {
            if (num[i]==k) f[k][i]=i+1;
            else {
                if (!k || !f[k-1][i] || !f[k-1][f[k-1][i]]) f[k][i]=0;
                else f[k][i]=f[k-1][f[k-1][i]];
            }
            if (f[k][i]) ans=k;
        }
        f[k][n+1]=0;
    }
    return;
}

int main()
{
    input();
    scanf("%d\n",&n);
    int i;
    for (i=1;i<=n;i++) scanf("%d",&num[i]);
    work();
    printf("%d\n",ans);
    output();
    return 0;
}

 

AU

【Splitting the Field】

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

在平面内给一些点，然后用两个矩形把所有的点围起来，矩形不能有交集和重边，问最小面积。N≤50,000

首先，两个矩形必定会在最左最右最上最下。进一步可以发现，答案一定在左右分割或者上下分割两种情况里面。

然后就排好序，扫一边求以这个点为分界划出的两个区域的面积。

两个答案取最大值。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<algorithm>
#include<stack>
#include<cmath>
#include<vector>
#include<map>
#define LL long long
#define maxn 100000
using namespace std;

struct node{LL x,y;} point[maxn];
LL nmax[maxn],nmin[maxn];
int n;

int cmp1(node p1,node p2){return p1.y<p2.y;}
int cmp2(node p1,node p2){return p1.x<p2.x;}

void input() {freopen("1.in","r",stdin);freopen("1.out","w",stdout);}
void output() {fclose(stdin);fclose(stdout);}
int main()
{
    input(); 
    scanf("%d",&n);
    int i;
    for (i=0;i<n;i++) scanf("%lld %lld",&point[i].x,&point[i].y);
    LL ans=((LL)(1)<<62),now1,now2,now,big1,big2;
    //x
//    printf("\n");
    sort(point,point+n,cmp1);
    big1=point[n-1].y-point[0].y; 
    nmax[n-1]=point[n-1].x;
    nmin[n-1]=point[n-1].x;
    for (i=n-2;i>=0;i--) {
        nmax[i]=max(point[i].x,nmax[i+1]);
        nmin[i]=min(point[i].x,nmin[i+1]);
    }
//    for (i=0;i<n;i++) printf("%lld %lld %lld %lld\n",point[i].x,point[i].y,nmin[i],nmax[i]);
    now1=point[0].x;
    now2=point[0].x;
    for (i=1;i<n-1;i++) {
        now1=max(point[i].x,now1);
        now2=min(point[i].x,now2);
        if (point[i+1].y!=point[i].y) {
            now=(point[i].y-point[0].y)*(now1-now2)
                +(point[n-1].y-point[i+1].y)*(nmax[i+1]-nmin[i+1]);
            if (now<ans) ans=now;
        }
    }
//    printf("%lld %d\n",ans,ans1);
    //y
//    printf("\n");
    sort(point,point+n,cmp2);
    big2=point[n-1].x-point[0].x;
    nmax[n-1]=point[n-1].y;
    nmin[n-1]=point[n-1].y;
    for (i=n-2;i>=0;i--) {
        nmax[i]=max(point[i].y,nmax[i+1]);
        nmin[i]=min(point[i].y,nmin[i+1]);
    }
//    for (i=0;i<n;i++) printf("%lld %lld %lld %lld\n",point[i].x,point[i].y,nmin[i],nmax[i]);
    now1=point[0].y;
    now2=point[0].y;
    for (i=1;i<n-1;i++) {
        now1=max(point[i].y,now1);
        now2=min(point[i].y,now2);
        if (point[i+1].x!=point[i].x) {
            now=(point[i].x-point[0].x)*(now1-now2)
                +(point[n-1].x-point[i+1].x)*(nmax[i+1]-nmin[i+1]);
            if (now<ans) ans=now;
        }
    }
    
//    printf("%lld %d\n",ans,ans1);
    printf("%lld\n",big1*big2-ans);
    output();
    return 0;
 } 

 

【Closing the Farm】

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

给一个图，每次去掉一个点，问每次去掉后余下的图是否联通。

倒过来，变成每次加一个点看是否图联通。并查集做就可以。

（并查集加了启发式合并然并卵）

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<algorithm>
#include<stack>
#include<cmath>
#include<vector>
#include<map>
#define maxn 201000
#define maxm 401000
using namespace std;

int to[maxm],next[maxm],first[maxn],ans[maxn],fa[maxn],rank[maxn],chose[maxn],num[maxn],n,m,total=0;


void input() {freopen("closing.in","r",stdin);freopen("closing.out","w",stdout);}
void output() {fclose(stdin);fclose(stdout);}
void addedge(int j,int k)
{
    next[++total]=first[j];
    to[first[j]=total]=k;
}
int find(int x)
{
    if (fa[x]==x) return x;
    return fa[x]=find(fa[x]);
}

int main()
{
    input();
    int i,j,k;
    scanf("%d %d",&n,&m);
    memset(first,0,sizeof(first));
    for (i=1;i<=m;i++) {
        scanf("%d %d",&j,&k);
        addedge(j,k);
        addedge(k,j);
    }
    for (i=1;i<=n;i++) scanf("%d",&num[i]);
    int now=0;
    for (i=1;i<=n;i++) fa[i]=i;
    memset(chose,0,sizeof(chose));
    memset(ans,0,sizeof(ans));
    memset(rank,0,sizeof(rank));
    for (i=n;i;i--) {
        ++now;
        int x=num[i];
        chose[x]=1;
        rank[x]=1;
        for (j=first[x];j;j=next[j]) {
            int too=to[j];
            if (chose[too]) {
                int fy=find(too);
                int fx=find(x);
                if (fy!=fx) {
                    --now;
                    if (rank[fx]>rank[fy]) fa[fy]=fx;
                    else {
                        fa[fx]=fy;
                        if (rank[fx]==rank[fy]) rank[fy]++;
                    }
                }
            }
        }
        if (now==1) ans[i]=1;
    }
    for (i=1;i<=n;i++)
        printf(ans[i]?"YES\n":"NO\n");
    output();
    return 0;
}

 

【248】

262144的弱化版。n^3dp就可以了。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<algorithm>
#include<stack>
#include<cmath>
#include<vector>
#include<map>
#define maxn 300
using namespace std;

int f1[maxn][maxn],f2[maxn][maxn],f3[maxn][maxn],num[maxn],ans=0,n;

void input() {freopen("1.in","r",stdin);freopen("1.out","w",stdout);}
void output() {fclose(stdin);fclose(stdout);}
void dfs(int ll,int rr)
{
    if (f1[ll][rr]!=-1) return;
    if (ll==rr) {
        f1[ll][rr]=num[ll];
        f2[ll][rr]=num[ll];
        f3[ll][rr]=num[ll];
        ans=max(ans,f1[ll][rr]);
        ans=max(ans,f2[ll][rr]);
        ans=max(ans,f3[ll][rr]);
        return;
    }
    if (ll+1==rr) {
        if (num[ll]==num[rr]) {
            f1[ll][rr]=num[ll]+1;
            f2[ll][rr]=num[ll]+1;
            f3[ll][rr]=num[rr]+1;
        }
        else {
            f1[ll][rr]=0;
            f2[ll][rr]=num[ll];
            f3[ll][rr]=num[rr];
        }
        ans=max(ans,f1[ll][rr]);
        ans=max(ans,f2[ll][rr]);
        ans=max(ans,f3[ll][rr]);
        return;
    }
    f1[ll][rr]=0;
    for (int i=ll;i<rr;i++) {
        dfs(ll,i);
        dfs(i+1,rr);
        if (f1[ll][i]==f1[i+1][rr] && f1[ll][i]) f1[ll][rr]=f1[ll][i]+1;
        if (f2[ll][i]>f2[ll][rr]) f2[ll][rr]=f2[ll][i];
        if (f3[i+1][rr]>f3[ll][rr]) f3[ll][rr]=f3[i+1][rr];
        if (f3[ll][i]==f2[i+1][rr] && f2[i+1][rr]) 
            ans=max(ans,f2[i+1][rr]+1);
        ans=max(ans,f1[ll][rr]);
        ans=max(ans,f2[ll][rr]);
        ans=max(ans,f3[ll][rr]);
    }
} 
 
 
int main()
{
    input();
    scanf("%d",&n);
    int i,j;
    for (i=1;i<=n;i++) scanf("%d",&num[i]);
    for (i=1;i<=n;i++)
        for (j=i;j<=n;j++)
            f1[i][j]=-1;
    dfs(1,n);
//    for (i=1;i<=n;i++)
//        for (j=i;j<=n;j++)
//            printf("%d %d:%d %d %d\n",i,j,f1[i][j],f2[i][j],f3[i][j]);
    printf("%d\n",ans);
    output();
    return 0;
}

 

ag

【Field Reduction】

给一些在平面内的点，去掉三个点，问余下的点用矩形圈起来的最小面积。

直接爆搜，每次只可能去最上最下最左最右的点。4^3。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<algorithm>
#include<stack>
#include<cmath>
#include<vector>
#include<map>
#define maxn 600000
using namespace std;

struct node{int x,y,num;} point[maxn],po2[maxn];
int ans,n,numx[maxn],numy[maxn],chose[maxn],p[3];


int cmp1(node x,node y){return x.x<y.x;}
int cmp2(node x,node y){return x.y<y.y;}

void input() {freopen("1.in","r",stdin);freopen("1.out","w",stdout);}
void output() {fclose(stdin);fclose(stdout);}
void dfs(int x)
{
    int l1=-1,l2=n,r1=-1,r2=n;
        while (!chose[numx[++l1]]);
        while (!chose[numx[--l2]]);
        while (!chose[numy[++r1]]);
        while (!chose[numy[--r2]]);
    if (x==3) {
    //    for (int i=0;i<3;i++) printf("%d ",p[i]);printf("\n");
    //    printf("%d %d %d %d %d %d %d\n",numx[l1],numx[l2],numy[r1],numy[r2],
    //            (po2[numx[l2]].x-po2[numx[l1]].x),
    //            (po2[numy[r2]].y-po2[numy[r1]].y),
    //            (po2[numx[l2]].x-po2[numx[l1]].x)*(po2[numy[r2]].y-po2[numy[r1]].y));
        ans=min((po2[numx[l2]].x-po2[numx[l1]].x)*(po2[numy[r2]].y-po2[numy[r1]].y),ans);
        return;
    }
    chose[numx[l1]]=0;
    p[x]=numx[l1];
    dfs(x+1);
    chose[numx[l1]]=1;
    
    chose[numx[l2]]=0;
    p[x]=numx[l2];
    dfs(x+1);
    chose[numx[l2]]=1;
    
    chose[numy[r1]]=0;
    p[x]=numy[r1];
    dfs(x+1);
    chose[numy[r1]]=1;
    
    chose[numy[r2]]=0;
    p[x]=numy[r2];
    dfs(x+1);
    chose[numy[r2]]=1;
}

int main()
{
    input();
    scanf("%d",&n);
    int i;
    for (i=0;i<n;i++) {
        scanf("%d %d",&point[i].x,&point[i].y);
        point[i].num=i; 
        po2[i]=point[i];
    }
    sort(point,point+n,cmp1);
    for (i=0;i<n;i++) numx[i]=point[i].num;
    sort(point,point+n,cmp2);
    for (i=0;i<n;i++) numy[i]=point[i].num;
//    for (i=0;i<n;i++) printf("%d ",numx[i]);printf("\n");
//    for (i=0;i<n;i++) printf("%d ",numy[i]);printf("\n");
    ans=(1<<31)-1;
    for (i=0;i<n;i++) chose[i]=1; 
    dfs(0);
    printf("%d\n",ans);
    output();
    return 0;
}

 

【Diamond Collector】

给一些数，从数中选取出两个集合，要求集合内的数差不超过给定值，问两个集合数字个数和最大？

队列就可以啦，出队为当前点和队头距离大于m。然后每次的答案为之前的最大值和当前队里面数的个数。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<algorithm>
#include<stack>
#include<cmath>
#include<vector>
#include<map>
#define LL long long
#define maxn 55000
using namespace std;

int f[maxn],n,ans,num[maxn],m;

void input() {freopen("1.in","r",stdin);freopen("1.out","w",stdout);}
void output() {fclose(stdin);fclose(stdout);}
int main()
{
    input();
    scanf("%d %d",&n,&m);
    int i;
    for (i=0;i<n;i++) scanf("%d",&num[i]);
    sort(num,num+n);
    int l=0,r=-1,ans=0,now=0;
    while (++r<n) {
        while (num[r]-num[l]>m) {
            if (f[l]>ans) ans=f[l];
            l++;
        }
        f[r]=r-l+1;
    //    printf("%d %d %d %d %d\n",l,r,num[r],f[r],ans);
        if (now<ans+f[r]) now=ans+f[r];
    }
//    for (i=0;i<n;i++) printf("%d %d %d\n",i,num[i],f[i]);
    printf("%d\n",now);
    output();
    return 0;
} 
 

 

【Closing the Farm】同上