BZOJ4152 AMPPZ2014 The Captain(最短路)

  事实上每次走到横坐标或纵坐标最接近的点一定可以取得最优方案。于是这样连边跑最短路就可以了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 200010
int n,p[N],d[N],t=0;
bool flag[N];
struct point{int x,y,i;
}a[N];
struct data{int to,nxt,len;
}edge[N<<2];
struct data2
{
    int x,d;
    bool operator <(const data2&a) const
    {
        return d>a.d;
    }
};
priority_queue<data2> q;
void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
void dijkstra()
{
    while (!q.empty()) q.pop();
    for (int i=2;i<=n;i++) d[i]=1000000001;q.push((data2){1,0});
    memset(flag,0,sizeof(flag));
    for (int i=1;i<=n;i++)
    {
        while (!q.empty()&&flag[q.top().x]) q.pop();
        if (q.empty()) break;
        data2 v=q.top();q.pop();
        flag[v.x]=1;
        for (int j=p[v.x];j;j=edge[j].nxt)
        if (v.d+edge[j].len<d[edge[j].to])
        {
            d[edge[j].to]=v.d+edge[j].len;
            q.push((data2){edge[j].to,d[edge[j].to]});
        }
    }
}
bool cmp1(const point&a,const point&b)
{
    return a.x<b.x;
}
bool cmp2(const point&a,const point&b)
{
    return a.y<b.y;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4152.in","r",stdin);
    freopen("bzoj4152.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i].x=read(),a[i].y=read(),a[i].i=i;
    sort(a+1,a+n+1,cmp1);
    for (int i=1;i<n;i++)
    addedge(a[i].i,a[i+1].i,min(abs(a[i].x-a[i+1].x),abs(a[i].y-a[i+1].y))),
    addedge(a[i+1].i,a[i].i,min(abs(a[i].x-a[i+1].x),abs(a[i].y-a[i+1].y)));
    sort(a+1,a+n+1,cmp2);
    for (int i=1;i<n;i++)
    addedge(a[i].i,a[i+1].i,min(abs(a[i].x-a[i+1].x),abs(a[i].y-a[i+1].y))),
    addedge(a[i+1].i,a[i].i,min(abs(a[i].x-a[i+1].x),abs(a[i].y-a[i+1].y)));
    dijkstra();
    cout<<d[n];
    return 0;
}

 

posted @ 2018-10-22 01:04  Gloid  阅读(161)  评论(0编辑  收藏  举报