[BZOJ4152]4152: [AMPPZ2014]The Captain

[BZOJ4152]4152: [AMPPZ2014]The Captain

题意

给定平面上的n个点，定义(x1,y1)到(x2,y2)的费用为min(|x1-x2|,|y1-y2|)，求从1号点走到n号点的最小费用。

题解

以X坐标排一遍序，将相邻的点连边，边权为X坐标的差

对Y再做一遍同样的操作

跑最短路即可

正确性显然

#include<bits/stdc++.h>
 
using namespace std;
 
const int MAXN = 200000 + 10;
 
inline int read()
{
    int f=1,x=0;
    char ch;
    do
    {
        ch=getchar();
        if(ch=='-') f=-1;
    }while(ch<'0'||ch>'9');
    do
    {
        x=(x<<3)+(x<<1)+ch-'0';
        ch=getchar();
    }while(ch>='0'&&ch<='9');
    return f*x;
} 
 
int n;
struct Query
{
    int x;
    int y;
    int id;
}a[MAXN];
 
inline bool cmp(Query a1,Query a2)
{
    return a1.x<a2.x;
}
 
inline bool cmp1(Query a1,Query a2)
{
    return a1.y<a2.y;
} 
 
struct node
{
    int id;
    long long dis;
    friend bool operator < (node a1,node a2)
    {
        return a1.dis>a2.dis;
    }
};
long long dis[MAXN];
 
priority_queue<node>q; 
 
struct Edge
{
    int to;
    int next;
    int val;
}g[MAXN*4];
 
int head[MAXN],cnt;
 
inline void addedge(int a,int b,int c)
{
    ++cnt;g[cnt].to=b;g[cnt].next=head[a];g[cnt].val=c;head[a]=cnt;
}
 
int main()
{
    n=read();
    for(int i=1;i<=n;i++) a[i].x=read(),a[i].y=read(),a[i].id=i;
    sort(a+1,a+n+1,cmp);
    for(int i=1;i<n;i++)
    {
        addedge(a[i].id,a[i+1].id,a[i+1].x-a[i].x);
        addedge(a[i+1].id,a[i].id,a[i+1].x-a[i].x);
    }
    sort(a+1,a+n+1,cmp1);
    for(int i=1;i<n;i++)
    {
        addedge(a[i].id,a[i+1].id,a[i+1].y-a[i].y);
        addedge(a[i+1].id,a[i].id,a[i+1].y-a[i].y);
    }
    for(int i=2;i<=n;i++) dis[i]=(1<<30);
    q.push(node{1,0});
    while(q.size())
    {
        node qu=q.top();q.pop();
        int u=qu.id;
        if(qu.dis!=dis[u]) continue;
        for(int i=head[u];i;i=g[i].next)
        {
            int v=g[i].to;
            if(dis[v]>dis[u]+g[i].val)
            {
                dis[v]=dis[u]+g[i].val;
                q.push(node{v,dis[v]});
            }
        }
    }
    cout<<dis[n]<<endl;
    return 0;
}