【HDU】5817 Ice Walls【极角排序+树状数组+最短路】

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题目链接：Ice Walls

题目大意：给出平面上一些与坐标轴平行且严格不相交的线段，穿过线段额外花费k时间，平面上移动一单位欧式距离花费1，求s到t的最短路。
题目分析：横竖线段分别考虑，先考虑竖线。枚举顶点，求出点和线段到顶点的x轴距离，排序，然后双指针，线段覆盖的是一个极角区间，因此可以树状数组求贡献。横线同理。

#include <bits/stdc++.h>using namespace std ;typedef long long LL ;typedef pair < double , int > pii ;#define clr( a , x ) memset ( a , x , sizeof a )const int MAXN = 1005 ;const double eps = 1e-8 ;const double pi = acos ( -1.0 ) ;struct Point {    int x , y ;    Point () {}    Point ( int x , int y ) : x ( x ) , y ( y ) {}    Point operator - ( const Point& a ) const {        return Point ( x - a.x , y - a.y ) ;    }    Point operator + ( const Point& a ) const {        return Point ( x + a.x , y + a.y ) ;    }    void input () {        scanf ( "%d%d" , &x , &y ) ;    }    double angle () {        return atan2 ( y , x ) ;    }    double dis () {        return sqrt ( 1.0 * x * x + 1.0 * y * y ) ;    }} ;struct Line {    int u , v ;} ;struct Node {    int dis , idx ;    bool operator < ( const Node& a ) const {        return dis < a.dis ;    }} ;Point p[MAXN] ;Line seg[MAXN] ;vector < pii > G ;Node a[MAXN << 1] ;int n , m , K , s , t ;int c[MAXN] ;double dis[MAXN][MAXN] ;double d[MAXN] ;int vis[MAXN] ;int rnk[MAXN] ;void add ( int x , int n , int v ) {    for ( int i = x ; i ; i -= i & -i ) {        c[i] += v ;    }}int sum ( int x , int n ) {    int ans = 0 ;    for ( int i = x ; i <= n ; i += i & -i ) {        ans += c[i] ;    }    return ans ;}int lfind ( double x , int l = 0 , int r = G.size () - 1 ) {    while ( l < r ) {        int m = l + r >> 1 ;        if ( x + eps < G[m].first ) r = m ;        else l = m + 1 ;    }    if ( x + eps > G[l].first ) ++ l ;    return l + 1 ;}int rfind ( double x , int l = 0 , int r = G.size () - 1 ) {    while ( l < r ) {        int m = l + r + 1 >> 1 ;        if ( G[m].first + eps < x ) l = m ;        else r = m - 1 ;    }    if ( G[l].first + eps > x ) -- l ;    return l + 1 ;}void deal ( int o , int cnt ) {    int num = G.size () ;    sort ( G.begin () , G.end () ) ;    sort ( a , a + cnt ) ;    for ( int i = 1 ; i <= num ; ++ i ) {        rnk[G[i - 1].second] = i ;        c[i] = 0 ;    }    for ( int i = 0 ; i < cnt ; ++ i ) {        int idx = a[i].idx ;        if ( idx < 0 ) {            int u = seg[-idx].u ;            int v = seg[-idx].v ;            double angl = ( p[u] - p[o] ).angle () ;            double angr = ( p[v] - p[o] ).angle () ;            if ( angl > angr ) swap ( angl , angr ) ;            int l = lfind ( angl ) ;            int r = rfind ( angr ) ;            add ( r , num , 1 ) ;            add ( l - 1 , num , -1 ) ;        } else {            LL t = 1LL * K * sum ( rnk[idx] , num ) ;            dis[o][idx] += t ;            dis[idx][o] += t ;        }    }}void dij ( int s ) {    for ( int i = 1 ; i <= m ; ++ i ) {        d[i] = dis[s][i] ;        vis[i] = 0 ;    }    d[s] = 0 ;    vis[s] = 1 ;    for ( int i = 1 ; i < m ; ++ i ) {        double minv = 1e60 ;        int u = s ;        for ( int j = 1 ; j <= m ; ++ j ) {            if ( !vis[j] && d[j] < minv ) {                u = j ;                minv = d[j] ;            }        }        vis[u] = 1 ;        for ( int j = 1 ; j <= m ; ++ j ) {            if ( !vis[j] && d[u] + dis[u][j] < d[j] ) {                d[j] = d[u] + dis[u][j] ;            }        }    }}void solve () {    m = 0 ;    scanf ( "%d%d" , &n , &K ) ;    for ( int i = 1 ; i <= n ; ++ i ) {        p[++ m].input () ;        seg[i].u = m ;        p[++ m].input () ;        seg[i].v = m ;    }    p[++ m].input () ;    s = m ;    p[++ m].input () ;    t = m ;    for ( int i = 1 ; i <= m ; ++ i ) {        for ( int j = i + 1 ; j <= m ; ++ j ) {            dis[i][j] = dis[j][i] = ( p[i] - p[j] ).dis () ;        }    }    for ( int i = 1 ; i <= m ; ++ i ) {        int cnt = 0 ;        G.clear () ;        for ( int j = 1 ; j <= m ; ++ j ) if ( j != i ) {            if ( p[j].x > p[i].x ) {                G.push_back ( pii ( ( p[j] - p[i] ).angle () , j ) ) ;                a[cnt].dis = p[j].x - p[i].x ;                a[cnt ++].idx = j ;            }        }        for ( int j = 1 ; j <= n ; ++ j ) {            int x = p[seg[j].u].x ;            if ( x == p[seg[j].v].x && x > p[i].x ) {                a[cnt].dis = x - p[i].x ;                a[cnt ++].idx = -j ;            }        }        deal ( i , cnt ) ;        cnt = 0 ;        G.clear () ;        for ( int j = 1 ; j <= m ; ++ j ) if ( j != i ) {            if ( p[j].y > p[i].y ) {                G.push_back ( pii ( ( p[j] - p[i] ).angle () , j ) ) ;                a[cnt].dis = p[j].y - p[i].y ;                a[cnt ++].idx = j ;            }        }        for ( int j = 1 ; j <= n ; ++ j ) {            int y = p[seg[j].u].y ;            if ( y == p[seg[j].v].y && y > p[i].y ) {                a[cnt].dis = y - p[i].y ;                a[cnt ++].idx = -j ;            }        }        deal ( i , cnt ) ;    }    dij ( s ) ;    printf ( "%.10f\n" , d[t] ) ;}int main () {    int T ;    scanf ( "%d" , &T ) ;    for ( int i = 1 ; i <= T ; ++ i ) {        solve () ;    }    return 0 ;}

压缩后代码：

#include <bits/stdc++.h>using namespace std;typedef long long LL;typedef pair<double,int>pii;#define clr(a,x) memset(a,x,sizeof a)const int MAXN=1005;const double eps=1e-8;const double pi=acos(-1.0);struct Point{    int x,y;    Point(){}    Point(int x,int y):x(x),y(y){}    Point operator - (const Point &a)const{return Point (x-a.x,y-a.y);}    Point operator + (const Point &a)const{return Point (x+a.x,y+a.y);}    void input(){scanf("%d%d",&x,&y);}    double angle(){return atan2(y,x);}    double dis(){return sqrt(1.0*x*x+1.0*y*y);}}p[MAXN];struct Line{    int u,v;}seg[MAXN];struct Node{    int dis,idx;    Node(){}    Node(int d,int i):dis(d),idx(i){}    bool operator < (const Node&a)const{return dis<a.dis;}}a[MAXN<<1];pii G[MAXN];int n,m,K,s,t,c[MAXN],vis[MAXN],rnk[MAXN],num;double dis[MAXN][MAXN],d[MAXN];void add(int x,int n,int v){    for(int i=x;i;i-=i&-i)c[i]+=v;}int sum(int x,int n,int ans=0){    for(int i=x;i<=n;i+=i&-i)ans+=c[i];    return ans;}int lfind(double x,int l=0,int r=num-1){    for(int m=l+r>>1;l<r;m=l+r>>1)x+eps<G[m].first?r=m:l=m+1;    if(x+eps>G[l].first)++l;    return l+1;}int rfind(double x,int l=0,int r=num-1){    for(int m=l+r+1>>1;l<r;m=l+r+1>>1)G[m].first+eps<x?l=m:r=m-1;    if(G[l].first+eps>x)--l;    return l+1;}void deal(int o,int cnt){    int i,idx;    sort(G,G+num);    sort(a,a+cnt);    for(i=1;i<=num;++i)rnk[G[i-1].second]=i,c[i]=0;    for(i=0,idx=a[i].idx;i<cnt;++i,idx=a[i].idx){        if(idx<0){            int u=seg[-idx].u,v=seg[-idx].v;            double angl=(p[u]-p[o]).angle();            double angr=(p[v]-p[o]).angle();            if(angl>angr)swap(angl,angr);            int l=lfind(angl),r=rfind(angr);            add(r,num,1);            add(l-1,num,-1);        }else{            LL t=1LL*K*sum(rnk[idx],num);            dis[o][idx]+=t;            dis[idx][o]+=t;        }    }}void dij(int s){    int i,j,u;    for(i=1;i<=m;++i)d[i]=dis[s][i],vis[i]=0;    for(d[s]=0,vis[s]=1,i=1,j,u;i<m;++i){        double minv=1e60;        for(u=s,j=1;j<=m;++j)if(!vis[j]&&d[j]<minv)minv=d[u=j];        for(vis[u]=1,j=1;j<=m;++j)if(!vis[j]&&d[u]+dis[u][j]<d[j])d[j]=d[u]+dis[u][j];    }}void solve(){    m=0;    scanf("%d%d",&n,&K);    for(int i=1;i<=n;++i)p[seg[i].u=++m].input(),p[seg[i].v=++m].input();    p[s=++m].input(),p[t=++m].input();    for(int i=1;i<=m;++i)for(int j=i+1;j<=m;++j)dis[i][j]=dis[j][i]=(p[i]-p[j]).dis();    for(int i=1,j,cnt;i<=m;++i){        for(num=0,cnt=0,j=1;j<=m;++j)if(j!=i&&p[j].x>p[i].x){            G[num++]=pii((p[j]-p[i]).angle(),j);            a[cnt++]=Node(p[j].x-p[i].x,j);        }        for(j=1;j<=n;++j){            int x=p[seg[j].u].x;            if(x==p[seg[j].v].x&&x>p[i].x)a[cnt++]=Node(x-p[i].x,-j);        }        deal(i,cnt);        for(num=0,cnt=0,j=1;j<=m;++j)if(j!=i&&p[j].y>p[i].y){            G[num++]=pii((p[j]-p[i]).angle(),j);            a[cnt++]=Node(p[j].y-p[i].y,j);        }        for(j=1;j<=n;++j){            int y=p[seg[j].u].y;            if(y==p[seg[j].v].y&&y>p[i].y)a[cnt++]=Node(y-p[i].y,-j);        }        deal(i,cnt);    }    dij(s);    printf("%.10f\n",d[t]);}int main(){    int T;    scanf("%d",&T);    for(int i=1;i<=T;++i)solve();    return 0;}

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