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  • 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
    ";
    #else
        const char LL[]="%lld
    ";
    #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;
    }
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  • 原文地址:https://www.cnblogs.com/Gloid/p/9827815.html
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