标签：contain amp pre rom not etc sizeof positive -o
Bahiyyah has a convex polygon with nn vertices P0,P1,?,Pn−1P0,P1,?,Pn−1 in the counterclockwise order. Two vertices with consecutive indexes are adjacent, and besides, P0P0 and Pn−1Pn−1 are adjacent. She also assigns a point QQ inside the polygon which may appear on the border.
Now, Bahiyyah decides to roll the polygon along a straight line and calculate the length of the trajectory (or track) of point QQ.
To help clarify, we suppose Pn=P0,Pn+1=P1Pn=P0,Pn+1=P1 and assume the edge between P0P0 and P1P1 is lying on the line at first. At that point when the edge between Pi−1Pi−1 and PiPi lies on the line, Bahiyyah rolls the polygon forward rotating the polygon along the vertex PiPi until the next edge (which is between PiPi and Pi+1Pi+1) meets the line. She will stop the rolling when the edge between PnPn and Pn+1Pn+1 (which is same as the edge between P0P0 and P1P1) meets the line again.
The input contains several test cases, and the first line is a positive integer TT indicating the number of test cases which is up to 5050.
For each test case, the first line contains an integer n (3≤n≤50)n (3≤n≤50) indicating the number of vertices of the given convex polygon. Following nn lines describe vertices of the polygon in the counterclockwise order. The ii-th line of them contains two integers xi−1xi−1 and yi−1yi−1, which are the coordinates of point Pi−1Pi−1. The last line contains two integers xQxQ and yQyQ, which are the coordinates of point QQ.
We guarantee that all coordinates are in the range of −103−103 to 103103, and point QQ is located inside the polygon or lies on its border.
For each test case, output a line containing Case #x: y, where x is the test case number starting from 11, and y is the length of the trajectory of the point QQ rounded to 33 places. We guarantee that 44-th place after the decimal point in the precise answer would not be 44 or 55.
4
4
0 0
2 0
2 2
0 2
1 1
3
0 0
2 1
1 2
1 1
5
0 0
1 0
2 2
1 3
-1 2
0 0
6
0 0
3 0
4 1
2 2
1 2
-1 1
1 0
Case #1: 8.886
Case #2: 7.318
Case #3: 12.102
Case #4: 14.537
分析：暴力附加两个数据，dp[0] = dp[n] , dp[n+1]=dp[1]，按余弦定理计算就能AC。
#include <iostream> #include <algorithm> #include <cstdio> #include <string> #include <cstring> #include <cstdlib> #include <map> #include <vector> #include <set> #include <queue> #include <stack> #include <cmath> using namespace std; #define pq priority_queue<int> #define pql priority_queue<ll> #define pqn priority_queue<node> #define v vector<int> #define vl vector<ll> #define lson rt<<1, l, m #define rson rt<<1|1, m+1, r #define read(x) scanf("%d",&x) #define lread(x) scanf("%lld",&x); #define pt(x) printf("%d\n",(x)) #define yes printf("YES\n"); #define no printf("NO\n"); #define gcd __gcd #define cn cin>> #define ct cout<< #define ed <<endl; #define ok return 0 ; #define over cout<<endl; #define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++) #define input(k) for (int i = 1; i <= (int)(k); i++) {cin>>a[i] ; } #define mem(s,t) memset(s,t,sizeof(s)) #define re return 0; #define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL); #define mod(x) ((x)%9973) #define test cout<<" ++++++ "<<endl; typedef long long ll; const int INT=1e6+5; const int maxn=10000+5; const int len = 1e5+5; typedef struct node { int x,y,ankle,T; }node; node dp[len]; double getlen(node xx,node yy) { return ( (xx.x-yy.x)*(xx.x-yy.x) +(xx.y-yy.y)*(xx.y-yy.y) ); } //计算两点间距离 int main() { TLE; int t,n,k; cin>>t; for(int kk=1;kk<=t;kk++) { double sum=0; cin>>n; rep(1,n) cin>>dp[i].x>>dp[i].y; dp[0].x=dp[n].x; dp[0].y=dp[n].y; dp[n+1].x=dp[1].x; dp[n+1].y=dp[1].y; cin>>dp[n+2].x>>dp[n+2].y; rep(1,n) { double dis = getlen(dp[i],dp[n+2]); double d1 = getlen(dp[i],dp[i+1]); double d2 = getlen(dp[i],dp[i-1]); double d3 = getlen(dp[i+1],dp[i-1]); sum+= sqrt( dis )*abs( acos(-1.0) - acos( ( d1+d2-d3)/(2*sqrt(d1)*sqrt(d2)) ) ); } printf("Case #%d: %.3lf\n",kk,sum); } ok; }
分析：数学签到题，（ 可以重点关注一下 小技巧 (i+1)%n 和 （i-1）%n ）
#include <iostream> #include <algorithm> #include <cstdio> #include <string> #include <cstring> #include <cstdlib> #include <map> #include <vector> #include <set> #include <queue> #include <stack> #include <cmath> using namespace std; #define pq priority_queue<int> #define pql priority_queue<ll> #define pqn priority_queue<node> #define v vector<int> #define vl vector<ll> #define lson rt<<1, l, m #define rson rt<<1|1, m+1, r #define read(x) scanf("%d",&x) #define lread(x) scanf("%lld",&x); #define pt(x) printf("%d\n",(x)) #define yes printf("YES\n"); #define no printf("NO\n"); #define gcd __gcd #define cn cin>> #define ct cout<< #define ed <<endl; #define ok return 0 ; #define over cout<<endl; #define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++) #define input(k) for (int i = 1; i <= (int)(k); i++) {cin>>a[i] ; } #define mem(s,t) memset(s,t,sizeof(s)) #define re return 0; #define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL); #define mod(x) ((x)%9973) #define test cout<<" ++++++ "<<endl; typedef long long ll; const int INT=1e6+5; const int maxn=10000+5; const int len = 1e5+5; typedef struct node { int x,y,ankle,T; }node; node dp[len]; double getlen(node xx,node yy) { return ( (xx.x-yy.x)*(xx.x-yy.x) +(xx.y-yy.y)*(xx.y-yy.y) ); } //计算两点间距离平方 int main() { TLE; int t,n,k; cin>>t; for(int kk=1;kk<=t;kk++) { double sum=0; cin>>n; rep(0,n-1) cin>>dp[i].x>>dp[i].y; cin>>dp[n+2].x>>dp[n+2].y; rep(0,n-1) { double dis = getlen(dp[i],dp[n+2]); double d1 = getlen(dp[i],dp[(i+1)%n]); double d2 = getlen(dp[i],dp[(i-1+n)%n]); double d3 = getlen(dp[(i+1)%n],dp[(i-1+n)%n]); sum+= sqrt( dis )*abs( acos(-1.0) - acos( ( d1+d2-d3)/(2*sqrt(d1)*sqrt(d2)) ) ); } printf("Case #%d: %.3lf\n",kk,sum); } ok; }
标签：contain amp pre rom not etc sizeof positive -o
原文地址：https://www.cnblogs.com/Shallow-dream/p/11441375.html
