[USACO 2016 Dec Gold] Tutorial

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A:

贪心从小到大插入，用并查集维护连通性

#include <bits/stdc++.h>

using namespace std;
#define X first
#define Y second
typedef double db;
typedef long long ll;
typedef pair<int,int> P;
const int MAXN=1e3+;
int n,tot,cnt,f[MAXN];P dat[MAXN];
struct edge{int x,y;ll w;}e[MAXN*MAXN];

ll sqr(int x){return 1ll*x*x;}
bool cmp(edge x,edge y){return x.w<y.w;}
int find(int x){return f[x]==x?x:f[x]=find(f[x]);}

int main()
{
    scanf("%d",&n);
    for(int i=;i<=n;i++)
        scanf("%d%d",&dat[i].X,&dat[i].Y),f[i]=i;
    for(int i=;i<n;i++)
        for(int j=i+;j<=n;j++)
            e[++tot]={i,j,sqr(dat[i].X-dat[j].X)+sqr(dat[i].Y-dat[j].Y)};
    sort(e+,e+tot+,cmp);

    cnt=n;
    for(int i=;i<=tot;i++)
    {
        int px=find(e[i].x),py=find(e[i].y);
        if(px!=py) f[px]=py,cnt--;
        if(cnt==) return printf("%lld",e[i].w),;
    }
    return ;
}

Problem A

B:

$dp[1...n][1...m][0/1]$，其中0/1表示当前在哪一边

#include <bits/stdc++.h>

using namespace std;
#define X first
#define Y second
typedef double db;
typedef long long ll;
typedef pair<int,int> P;
const int MAXN=1e3+,INF=0x3f3f3f3f;
ll dp[MAXN][MAXN][];
int n,m;P dat[MAXN*];

ll sqr(int x){return 1ll*x*x;}
ll dist(int a,int b){return sqr(dat[a].X-dat[b].X)+sqr(dat[a].Y-dat[b].Y);}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=;i<=n+m;i++)
        scanf("%d%d",&dat[i].X,&dat[i].Y);

    memset(dp,0x3f,sizeof(dp));
    dp[][][]=;
    for(int i=;i<=n;i++)
        for(int j=;j<=m;j++)
            for(int k=;k<;k++)
            {
                if(i) dp[i+][j][]=min(dp[i+][j][],dp[i][j][]+dist(i,i+));
                if(j) dp[i+][j][]=min(dp[i+][j][],dp[i][j][]+dist(n+j,i+));
                if(i) dp[i][j+][]=min(dp[i][j+][],dp[i][j][]+dist(i,n+j+));
                if(j) dp[i][j+][]=min(dp[i][j+][],dp[i][j][]+dist(n+j,n+j+));
            }
    printf("%lld",dp[n][m][]);
    return ;
}

Problem B

C:

可以明显发现是最短路模型，但如果将一个点能不转弯走到的所有点的边都连上明显$TLE$

那么在跑最短路时多记录一维当前方向即可，每次转移判断是否要转向来决定是否增加代价

这样只要与不穿过其他点就能到达的点连边就行了

#include <bits/stdc++.h>

using namespace std;
#define X first
#define Y second
typedef double db;
typedef long long ll;
typedef pair<int,int> P;
const int MAXN=1e5+,INF=0x3f3f3f3f;
int n,x,y,h1[MAXN],h2[MAXN],dist[MAXN][],tot;
struct edge{int nxt,to;}e1[MAXN<<],e2[MAXN<<];
struct node{int x,y,id;}tmp[MAXN],dat[MAXN];
struct Queue{int w,x,dir;};

bool cmp1(node a,node b){return a.x<b.x;}
bool cmp2(node a,node b){return a.y<b.y;}
bool operator < (Queue a,Queue b){return a.w>b.w;}
void add1(int x,int y)
{
    e1[++tot]={h1[x],y};h1[x]=tot;
    e1[++tot]={h1[y],x};h1[y]=tot;
}
void add2(int x,int y)
{
    e2[++tot]={h2[x],y};h2[x]=tot;
    e2[++tot]={h2[y],x};h2[y]=tot;
}
priority_queue<Queue> q;

int main()
{
    scanf("%d",&n);n+=;
    scanf("%d%d",&x,&y);dat[]=tmp[]={x,y,};
    scanf("%d%d",&x,&y);dat[n]=tmp[n]={x,y,n};
    for(int i=;i<n;i++)
        scanf("%d%d",&dat[i].x,&dat[i].y),dat[i].id=i,tmp[i]=dat[i];

    sort(tmp+,tmp+n+,cmp1);
    for(int i=;i<n;i++)
        if(tmp[i].x==tmp[i+].x)
            add1(tmp[i].id,tmp[i+].id);
    sort(tmp+,tmp+n+,cmp2);
    for(int i=;i<n;i++)
        if(tmp[i].y==tmp[i+].y)
            add2(tmp[i].id,tmp[i+].id);

    memset(dist,0x3f,sizeof(dist));
    dist[][]=dist[][]=;
    q.push(Queue{,,});q.push(Queue{,,});
    while(!q.empty())
    {
        Queue t=q.top();q.pop();
        if(dist[t.x][t.dir]<t.w) continue;

        for(int i=h1[t.x];i;i=e1[i].nxt)
            if(dist[e1[i].to][]>dist[t.x][t.dir]+(t.dir!=))
                dist[e1[i].to][]=dist[t.x][t.dir]+(t.dir!=),q.push(Queue{dist[e1[i].to][],e1[i].to,});
        for(int i=h2[t.x];i;i=e2[i].nxt)
            if(dist[e2[i].to][]>dist[t.x][t.dir]+(t.dir!=))
                dist[e2[i].to][]=dist[t.x][t.dir]+(t.dir!=),q.push(Queue{dist[e2[i].to][],e2[i].to,});
    }
    int res=min(dist[n][],dist[n][]);
    printf("%d",res==INF?-:res);
    return ;
}

Problem C

不过看了题解发现由于：

1、同一行/列可以直达任一点

2、每次代价增加必然是行列间转化

从而也可以仅考虑坐标，离散化后对于点$(x,y)$将$x$行和$y$列连边即可