【DP】 HDOJ 5033 Building

3018 단어 HDU
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#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 400005
#define maxm 300005
#define eps 1e-10
#define mod 10000007
#define INF 1e9
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='
')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();} LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);} // head struct point { double x, y; int kk; point(double x = 0, double y = 0) : x(x), y(y) {} }; typedef point vec; vec operator - (vec a, vec b){return vec(a.x - b.x, a.y - b.y);} vec operator + (vec a, vec b){return vec(a.x + b.x, a.y + b.y);} vec operator / (vec a, double p){return vec(a.x / p, a.y / p);} vec operator * (vec a, double p){return vec(a.x * p, a.y * p);} int dcmp(double p){return (p > eps) - (p < eps);} double cross(point a, point b){return a.x * b.y - a.y * b.x;} point p[maxn], s[maxn]; double res[maxn]; int n, m; int cmp1(point a, point b){return a.x < b.x;} int cmp2(point a, point b){return a.x > b.x;} void read(void) { scanf("%d", &n); for(int i = 1; i <= n; i++) scanf("%lf%lf", &p[i].x, &p[i].y), p[i].kk = -1; scanf("%d", &m); for(int i = n+1; i <= n+m; i++) scanf("%lf", &p[i].x), p[i].y = 0, p[i].kk = i-n; n += m; } void work(void) { int top = 0; sort(p+1, p+n+1, cmp1); for(int i = 0; i <= m; i++) res[i] = 0; for(int i = 1; i <= n; i++) { while(top > 1 && dcmp(cross(s[top-2] - p[i], s[top-1] - p[i])) > 0) top--; if(p[i].kk == -1) s[top++] = p[i]; else res[p[i].kk] += atan((p[i].x - s[top-1].x) / s[top-1].y) / acos(-1.0) * 180.0; } top = 0; sort(p+1, p+n+1, cmp2); for(int i = 1; i <= n; i++) { while(top > 1 && dcmp(cross(s[top-2] - p[i], s[top-1] - p[i])) < 0) top--; if(p[i].kk == -1) s[top++] = p[i]; else res[p[i].kk] += atan((s[top-1].x - p[i].x) / s[top-1].y) / acos(-1.0) * 180.0; } for(int i = 1; i <= m; i++) printf("%.10f
", res[i]); } int main(void) { int _, __ = 0; while(scanf("%d", &_)!=EOF) { while(_--) { read(); printf("Case #%d:
", ++__); work(); } } return 0; }

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