[国家集训队]飞飞侠

https://www.luogu.org/problemnew/show/P4473

仙人题啊仙人题

正解是什么并查集优化？？看不懂也不会写

读题发现，这题的问题在于暴力建边会GG,需要解决的就是一个点向一个连通块连边的问题

想到可以把连通块拆成一行一行，在一行内，要解决一个点向一个区间连边的问题

线段树优化连边，我们只建维护入边的线段树，对于一个区间找到对应线段树内节点连边，树内从父亲向儿子连边（边权当然是0）

nm个点，e=n^2mlogm = 16875000个边，这样跑一边最短路是O(eloge)，似乎能过？？？

然后就是我的zz时间：这个图不是无向的，因此x->y和y->x显然不同，得挨个点跑一边最短路才行...

我好菜qaq

2.0h

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath> 
#include<iostream>
using namespace std;
#define O(x) cout << #x << " " << x << endl;
#define O_(x) cout << #x << " " << x << "  ";
#define B cout << "breakpoint" << endl;
typedef long long ll;
typedef pair<ll,int> pii;
const ll inf = 1e14;
#define mp make_pair
inline int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') op = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        (ans *= 10) += ch - '0';
        ch  = getchar();
    }
    return ans * op;
}
const int maxn = 303;
const int N = 1e6 + 5;
int n,m;
int id[maxn][maxn];//id[i][j] : i行j列对应线段树中哪个节点 
int ls[N],rs[N],root[maxn],tot;
struct egde
{
  int to,next;
  ll cost;
}e[N * 20];
int fir[N],alloc;
void adde(int u,int v,int w)
{
    e[++alloc].next = fir[u];
    fir[u] = alloc;
    e[alloc].to = v;
    e[alloc].cost = w;
}
int build(int line,int i,int l,int r)
{
    if(l == r)
    {
        id[line][l] = i;
        return i;
    }
    int mid = l + r >> 1;
    ls[i] = build(line,++tot,l,mid);
    rs[i] = build(line,++tot,mid + 1,r);
    adde(i,ls[i],0); adde(i,rs[i],0);
    return i;
}
void add(int i,int l,int r,int u,int ql,int qr,ll w)//u->[ql,qr]
{
    if(l == ql && r == qr)
    {
        adde(u,i,w);
        return;
    }
    int mid = l + r >> 1;
    if(qr <= mid) add(ls[i],l,mid,u,ql,qr,w);
    else if(ql > mid) add(rs[i],mid + 1,r,u,ql,qr,w);
    else add(ls[i],l,mid,u,ql,mid,w),add(rs[i],mid + 1,r,u,mid + 1,qr,w); 
}
bool vis[N];
ll dis[N];
void dij(int s)
{
        for(int i = 1;i <= tot;i++) dis[i] = inf;
    memset(vis,0,sizeof(vis));
    dis[s] = 0;
    priority_queue<pii,vector<pii>,greater<pii> > p;
    p.push(mp(dis[s],s));
    while(p.size())
    {
        pii t = p.top();
        p.pop();
        int u = t.second;
        if(vis[u]) continue;
        vis[u] = 1;
        for(int i = fir[u];i;i = e[i].next)
        {
      int v = e[i].to;
      ll w = e[i].cost;
            if(dis[v] > dis[u] + w)
            {
                dis[v] = dis[u] + w;
                if(!vis[v]) p.push(mp(dis[v],v));
            }
        }
    }
}
int b[maxn][maxn];
ll a[maxn][maxn];
int main()
{
    n = read(),m = read();
    for(int i = 1;i <= n;i++) root[i] = build(i,++tot,1,m);
    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= m;j++)
            b[i][j] = read();//距离 
    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= m;j++)
          scanf("%lld",&a[i][j]);//费用
    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= m;j++)
        {
            for(int k = max(1,i - b[i][j]);k <= min(n,i + b[i][j]);k++)
            {
                int res = b[i][j] - abs(k - i);
                add(root[k],1,m,id[i][j],max(1,j - res),min(m,j + res),a[i][j]);
            }
        }
    int x1 = read(),y1 = read(),x2 = read(),y2 = read(),x3 = read(),y3 = read();
    int x = id[x1][y1],y = id[x2][y2],z = id[x3][y3];
    dij(x);
    ll xy = dis[y],xz = dis[z];
    dij(y);
    ll yz = dis[z],yx = dis[x];
    dij(z);
    ll zx = dis[x],zy = dis[y];
    ll ansx = yx + zx,ansy = xy + zy,ansz = xz + yz;
    if(ansx >= inf && ansy >= inf && ansz >= inf)
      {
        printf("NO
");
        return 0;
      }
    if(ansx <= ansy && ansx <= ansz) printf("X
%lld",ansx);
    else if(ansy <= ansx && ansy <= ansz) printf("Y
%lld",ansy);
    else printf("Z
%lld",ansz);
}