bzoj 3116 登顶计划题解
作者：互联网
/*
{
######################
# Author #
# Gary #
# 2021 #
######################
*/
#include<bits/stdc++.h>
#define rb(a,b,c) for(int a=b;a<=c;++a)
#define rl(a,b,c) for(int a=b;a>=c;--a)
#define LL long long
#define IT iterator
#define PB push_back
#define II(a,b) make_pair(a,b)
#define FIR first
#define SEC second
#define FREO freopen("check.out","w",stdout)
#define rep(a,b) for(int a=0;a<b;++a)
#define SRAND mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define random(a) rng()%a
#define ALL(a) a.begin(),a.end()
#define POB pop_back
#define ff fflush(stdout)
#define fastio ios::sync_with_stdio(false)
#define check_min(a,b) a=min(a,b)
#define check_max(a,b) a=max(a,b)
using namespace std;
//inline int read(){
// int x=0;
// char ch=getchar();
// while(ch<'0'||ch>'9'){
// ch=getchar();
// }
// while(ch>='0'&&ch<='9'){
// x=(x<<1)+(x<<3)+(ch^48);
// ch=getchar();
// }
// return x;
//}
const int INF=0x3f3f3f3f;
typedef pair<int,int> mp;
/*}
*/
struct vec{
long double x,y;
long double operator * (vec oth){
return x*oth.y-y*oth.x;
}
vec operator - (vec oth){
return vec{x-oth.x,y-oth.y};
}
vec operator + (vec oth){
return vec{x+oth.x,y+oth.y};
}
};
bool cmp(vec A,vec B){
return A.x<B.x;
}
const int MAXN=2e5+233;
const long double eps=1e-7;
int n;
vec a[MAXN];
int sz;
vec sta[MAXN];
int sta2[MAXN];
vector<vec> ip;
void cross(vec A,vec B,vec C,vec D){
long double k=(A.y-B.y)/(A.x-B.x);
long double b=A.y-A.x*k;
long double k2,b2;
k2=(C.y-D.y)/(C.x-D.x);
b2=C.y-C.x*k2;
long double x,y;
if(abs(k-k2)<eps) return;
if(C.x>D.x) swap(C,D);
x=(b2-b)/(k-k2);
y=x*k+b;
if(x+eps<D.x&&x-eps>C.x){
ip.PB(vec{x,y});
}
}
int le[MAXN],ri[MAXN],id[MAXN];
long double tar[MAXN];
bool equ(vec A,vec B){
return abs(A.x-B.x)<eps&&abs(A.y-B.y)<eps;
}
vector<int> g[MAXN];
long double dis[MAXN];
long double rest[MAXN];
long double get(int u,int v){
if(u>v) swap(u,v);
return dis[v]-dis[u];
}
void dfs(int now){
for(auto it:g[now]){
rest[it]=rest[now]+get(it,now);
dfs(it);
}
}
int main(){
scanf("%d",&n);
rb(i,1,n) scanf("%Lf%Lf",&a[i].x,&a[i].y);
sz=0;
rb(i,1,n) ip.PB(a[i]);
rb(i,1,n){
while(sz>=2&&(sta[sz-1]-sta[sz])*(a[i]-sta[sz])<eps) {
cross(sta[sz-1],sta[sz],a[i-1],a[i]);
--sz;
}
sta[++sz]=a[i];
}
sz=0;
rl(i,n,1){
while(sz>=2&&(sta[sz-1]-sta[sz])*(a[i]-sta[sz])>-eps) {
cross(sta[sz-1],sta[sz],a[i+1],a[i]);
--sz;
}
sta[++sz]=a[i];
}
sort(ALL(ip),cmp);
vector<vec> v;
vec last=vec{INF,INF};
for(auto it:ip){
if(equ(last,it)){
continue;
}
last=it;
v.PB(it);
}
int m=v.size();
int to=1;
rep(i,m){
while(to<=n&&a[to].x+eps<v[i].x)++to;
if(to<=n&&equ(a[to],v[i])) id[to]=i;
}
rep(i,m){
// printf("(%.6Lf %.6Lf)\n",v[i].x,v[i].y);
le[i]=ri[i]=i;
tar[i]=v[i].y;
}
sz=0;
rep(i,m){
while(sz>=2&&(sta[sz-1]-sta[sz])*(v[i]-sta[sz])<eps) {--sz;}
if(sz>=1&&sta[sz].y>v[i].y) le[i]=sta2[sz],tar[i]=sta[sz].y;
sta[++sz]=v[i];
sta2[sz]=i;
}
sz=0;
rl(i,m-1,0){
while(sz>=2&&(sta[sz-1]-sta[sz])*(v[i]-sta[sz])>-eps) {--sz;}
if(sz>=1&&sta[sz].y>v[i].y){
ri[i]=sta2[sz];
if(v[ri[i]].y>tar[i]) tar[i]=v[ri[i]].y;
// cout<<i<<endl;
}
sta[++sz]=v[i];
sta2[sz]=i;
}
int root=0;
rep(i,m){
if(ri[i]==le[i]) root=i;
}
stack<pair<long double ,int> > stk;
rep(i,m){
if(i==root) continue;
while(!stk.empty()&&stk.top().FIR<=tar[i]+eps){stk.pop();}
if(v[le[i]].y>v[ri[i]].y){
if(stk.empty()||stk.top().SEC<=le[i]) g[le[i]].PB(i);
else g[stk.top().SEC].PB(i);
}
stk.push(II(tar[i],i));
}
while(!stk.empty()) stk.pop();
rl(i,m-1,0){
if(i==root) continue;
while(!stk.empty()&&stk.top().FIR<=tar[i]+eps){stk.pop();}
if(v[ri[i]].y>=v[le[i]].y){
if(stk.empty()||stk.top().SEC>=ri[i]) g[ri[i]].PB(i);
else g[stk.top().SEC].PB(i);
}
stk.push(II(tar[i],i));
}
// rep(i,m){
// cout<<le[i]<<' '<<ri[i]<<' ';
// printf("%.6Lf\n",tar[i]);
// }
rb(i,1,m){
double X,Y;
X=v[i].x-v[i-1].x;
Y=v[i].y-v[i-1].y;
X=X*X;
Y=Y*Y;
dis[i]=sqrt(X+Y);
dis[i]+=dis[i-1];
}
rest[root]=0;
dfs(root);
rb(i,1,n){
printf("%.6Lf\n",rest[id[i]]);
}
return 0;
}
标签：oth,sz,sta,题解,long,double,vec,3116,bzoj 来源： https://www.cnblogs.com/gary-2005/p/14396906.html