USACO 2017 February Contest总结

T1 Why Did the Cow Cross the Road

题目大意：给定一个 $N\times N$ 的网格。穿过两个格子的交界处需要有一个花费，每走三个格子也会有一个花费，问从左上角走到右下角的最小花费。

思路：一眼看上去每三个格有一个花费，看上去不是很好处理，仔细一想。
想
想。
发现我们可以将每三步化成一步，看一个点走三步可以到哪，就连边，边权为三次穿越边界的权值+这个点的权值。

WA了N次。。。原因：反向边和正向边。。。边权不相等。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<map>
#include<vector>
#include<ctime>
#include<stack>
#include<cctype>
#include<set>
#define mp make_pair
#define pa pair<long long,long long>
#define INF 0x3f3f3f3f
#define inf 0x3f
#define fi first
#define se second
#define pb push_back
#define ll long long
#define ull unsigned long long

using namespace std;

inline ll read()
{
    long long f=1,sum=0;
    char c=getchar();
    while (!isdigit(c)){if (c=='-') f=-1;c=getchar();}
    while (isdigit(c)){sum=sum*10+c-'0';c=getchar();}
    return sum*f;
}
const int MAXN=10010;
struct edge
{
    int next,to;
    ll val;
};
edge e[MAXN*40];
int head[MAXN],cnt;
void addedge(int u,int v,ll w)
{
    e[++cnt].next=head[u];
    e[cnt].to=v;
    e[cnt].val=w;
    head[u]=cnt;
}
priority_queue <pa,vector<pa>,greater<pa> > q;
ll dis[MAXN],n;
bool visit[MAXN];
#define id(x,y) (x-1)*n+y
void Dijkstra()
{
    for (int i=1;i<=n*n;i++)
        dis[i]=1e15;
    dis[1]=0;
    q.push(make_pair(0,1));
    while (!q.empty())
    {
        int x=q.top().se;
        q.pop();
        if (visit[x]) continue;
        visit[x]=1;
        for (int i=head[x];i;i=e[i].next)
        {
            int v=e[i].to;
            if (dis[x]+e[i].val<dis[v])
                dis[v]=dis[x]+e[i].val,q.push(make_pair(dis[v],v));
        }
    }   
} 
int a[110][110];
const int dx[16]={3,0,0,-3,1,2,-1,-2,2,-1,-2,1,0,1,0,-1};
const int dy[16]={0,3,-3,0,2,1,-2,-1,-1,2,1,-2,1,0,-1,0};
int main()
{
    int cost;
    scanf("%d%d",&n,&cost);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
            scanf("%d",&a[i][j]);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        {
            for (int k=0;k<16;k++)
            {
                int nx=i+dx[k],ny=j+dy[k];
                if (nx<=0 || ny<=0 || nx>n || ny>n) continue;
                addedge(id(i,j),id(nx,ny),cost*3+a[nx][ny]);
            }
        }
    Dijkstra();
    ll ans=INF;
    for (int i=n-2;i<=n;i++)
        for (int j=n-2;j<=n;j++)
        {
            int dist=(n-i)+(n-j);
            if (dist>=3) continue;
            ll now=dist*cost+dis[id(i,j)];
            ans=min(ans,now);
        }
    cout<<ans;
    return 0;   
}

T2 Why Did the Cow Cross the Road II

题目链接

题目大意：左边一个 $1\sim N$ 的排列，右边一个 $1\sim N$ 的排列，现在可以让一个点向另一侧权值和他相差不超过4的点连边。问在边不交叉的情况下最多连几条边。

思路： $N\le 1000$ ，感觉不是DP就是网络流，先想了想网络流，感觉无法建模，然后考虑DP。
- $f[i][j]$ 表示左边前 $i$ 个点，右边前 $j$ 个点匹配不交叉的最大匹配数。
- 转移：向前转移（i向i+1转移）。考虑 $i+1$ 的匹配，应该找到 $j+1\sim N$ 所有与它权值不超过4的点。预处理位置即可。
- 复杂度 $O(n^2)$ ，转移常数 $9$ 。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<map>
#include<vector>
#include<ctime>
#include<stack>
#include<cctype>
#include<set>
#define mp make_pair
#define pa pair<int,int>
#define INF 0x3f3f3f3f
#define inf 0x3f
#define fi first
#define se second
#define pb push_back
#define ll long long
#define ull unsigned long long

using namespace std;

inline ll read()
{
    long long f=1,sum=0;
    char c=getchar();
    while (!isdigit(c)){if (c=='-') f=-1;c=getchar();}
    while (isdigit(c)){sum=sum*10+c-'0';c=getchar();}
    return sum*f;
}
const int MAXN=1010;
int a[MAXN],b[MAXN],vis1[MAXN],vis2[MAXN];
int f[MAXN][MAXN];
int main()
{
    freopen("nocross.in","r",stdin);
    freopen("nocross.out","w",stdout); 
    int n;
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        scanf("%d",&a[i]),vis1[a[i]]=i;
    for (int i=1;i<=n;i++)
        scanf("%d",&b[i]),vis2[b[i]]=i;
    int ans=0;
    for (int i=1;i<=n;i++)
        f[1][i]=(abs(a[1]-b[i])<=4);
    for (int i=1;i<=n;i++)
    {
        for (int j=0;j<=n;j++)
        {
            f[i][j]=max(f[i][j],max(f[i-1][j],f[i][j-1==0?j:j-1]));
            ans=max(ans,f[i][j]);
            for (int k=-4;k<=4;k++)
            {
                int val=a[i+1]+k;
                if (val<=0 || val>n || vis2[val]<=j) continue;
                f[i+1][vis2[val]]=max(f[i+1][vis2[val]],f[i][j]+1); 
            }
        }

    }
    cout<<ans;
    return 0;
}

T3 Why Did the Cow Cross the Road III

题目链接

题目大意：给定一些线段的左端点和右端点，求相交的线段条数。

思路：树状数组。
将所有线段按照左端点排序，每次查询线段区间内有几个右端点即可。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<map>
#include<vector>
#include<ctime>
#include<stack>
#include<cctype>
#include<set>
#define mp make_pair
#define pa pair<int,int>
#define INF 0x3f3f3f3f
#define inf 0x3f
#define fi first
#define se second
#define pb push_back
#define ll long long
#define ull unsigned long long

using namespace std;

inline ll read()
{
    long long f=1,sum=0;
    char c=getchar();
    while (!isdigit(c)){if (c=='-') f=-1;c=getchar();}
    while (isdigit(c)){sum=sum*10+c-'0';c=getchar();}
    return sum*f;
}
const int MAXN=50010;
struct node
{
    int l,r;
}a[MAXN];
int lowbit(int x){return x&(-x);}
int f[2*MAXN],n;
void update(int x)
{
    for (int i=x;i<=2*n;i+=lowbit(i))
        f[i]++;
}
int query(int x)
{
    int ans=0;
    for (int i=x;i;i-=lowbit(i))
        ans+=f[i];
    return ans;
}
bool visit[MAXN];
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=2*n;i++)
    {
        int x;
        scanf("%d",&x);
        if (visit[x])
            a[x].r=i;
        else
            a[x].l=i,visit[x]=1;
    }
    sort(a+1,a+1+n,[](node i,node j){return i.l<j.l;});
    int ans=0;
    for (int i=1;i<=n;i++)
    {
        ans+=query(a[i].r)-query(a[i].l-1);
        update(a[i].r);
    }
    cout<<ans;
    return 0;
}