BAPC2014 C&&hunnu11583:citadel Construction (geometry)

Test instructions

Give a series of points that require searching for up to 4 points to make up a polygon of the largest area

Ideas:

Obviously there are only two cases, either a triangle or a quadrilateral.

First of all, it is not difficult to think of the first to find out the most out of the point, is actually looking for convex bag

But there is no convex hull, so what to do?

We know that when the AB is multiplied with the AC vector to get ab*ac>=0, we can know that all of these conditions are fixed in the clockwise direction, so we can use this to find all the outer points.

After we get these points, we can figure out the answers.

#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <queue > #include <map> #include <set> #include <vector> #include <math.h> #include <bitset># Include <algorithm> #include <climits>using namespace std; #define LS 2*i#define rs 2*i+1#define up (i,x,y) for (i=x;i<=y;i++) #define DOWN (i,x,y) for (i=x;i>=y;i--) #define MEM (a,x) memset (A,x,sizeof (a)) #define W (a) while (a) #define GCD (A, B) __gcd (A, b) #define LL long long#define ULL unsigned long long#define N 1005#define INF 0x3f3f3f3f#define E  XP 1e-8#define rank rank1const int mod = 1000000007;struct point{int x, y;} a[n],s[n];int t,n;int ads (int a) {return A&LT;0?-A:A;} int mult (point a,point b,point c) {return (a.x-c.x) * (B.Y-C.Y)-(b.x-c.x) * (A.Y-C.Y);}    int CMP (point A,point B) {if (A.Y==B.Y) return a.x<b.x; return A.Y&LT;B.Y;}    int Set1 (point a[],int n,point s[]) {int i,j,k;    int top = 1;    Sort (a,a+n,cmp); if (n==0) RETUrn 0;    S[0] = a[0];    if (n==1) return 1;    S[1] = a[1];    if (n==2) return 2;    S[2] = a[2];            for (i = 2; i<n; i++)//First Look for the right outer layer {while (Top&&mult (a[i],s[top],s[top-1]) >=0)//new incoming in the clockwise direction, can replace the stored in the stack        top--;    S[++top] = A[i];    } int len = top;    S[++top] = a[n-2];         for (i = n-3; i>=0; i--)//Find the LEFT outer {while (Top!=len&&mult (a[i],s[top],s[top-1]) >=0) top--;    S[++top] = A[i]; } return top;}    void Solve () {int i,j,k,cnt,area;    for (int i=0; i<n; i++) scanf ("%d%d", &a[i].x,&a[i].y);    CNT = Set1 (a,n,s);    if (cnt<3) {printf ("0\n");        } else if (cnt==3) {area = Mult (S[2],s[1],s[0]);        Area = Ads (area);        if (area%2) printf ("%d.5\n", AREA/2);    else printf ("%d\n", AREA/2);        } else {area=-inf;        S[CNT] = s[0];            for (i = 0; i<cnt; i++) {int L = I,r = i+2; for (j = i+2; j<cnt; j + +)//with I,jFor quadrilateral Diagonal, both sides of the loop ensure that two triangles do not intersect to calculate the maximum area {while (ABS (Mult (s[l+1],s[j],s[i)) >abs (Mult (S[l],s[j],s[i]))) L                = (l+1)%cnt;                int S1 = ABS (Mult (s[l],s[j],s[i));                while (ABS (Mult (s[r+1],s[j],s[i))) >abs (Mult (S[r],s[j],s[i]))) R = (r+1)%cnt;                int s2 = ABS (Mult (s[r],s[j],s[i));            Area = max (AREA,S1+S2);        }} if (area%2) printf ("%d.5\n", AREA/2);    else printf ("%d\n", AREA/2);    }}int Main () {scanf ("%d", &t);        while (t--) {scanf ("%d", &n);    Solve (); } return 0;}

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BAPC2014 C&&hunnu11583:citadel Construction (geometry)