HDU3952 Fruit Ninja (geometry)

The problem is asking you to ask for a line that can pass through the most fruit (a point is counted).

It can be proved that it is possible to enumerate the lines consisting of two points.

Because if there is a line that crosses n fruits, it is necessary to translate some of it so that it passes through n but it can no longer be translated, so the line must be at some end of a certain fruit.

Again with this end, rotate this line, or through N, until it cannot be rotated (then the rotation may not pass through N), so that the line must have encountered another endpoint.

So just enumerate two endpoints.

There is only one fruit, and a special sentence.

#include <map> #include <set> #include <queue> #include <stack> #include <math.h> #include <time.h> #include <stdio.h> #include <stdlib.h> #include <iostream> #include <limits.h># include<string.h> #include <string> #include <algorithm> #define MID (x, y) >> 1 #define L ( x) (x << 1) #define R (x) (x << 1 | 1) using namespace Std;const int MAX = 35;struct point {int x, y;}; struct Polygon {point P[max]; int n,};p olygon g[max];int crossproduct (Point a,point b,point c) {return (c.x-a.x) * (b.y -A.Y)-(b.x-a.x) * (C.Y-A.Y);} BOOL S21_inst (Point s1,point s2,point l1,point L2) {return crossproduct (L1,L2,S1) *crossproduct (L1,L2,S2) <= 0;}    int solve (point a,point b,int n) {int ans=0;        for (int i=0;i<n;i++) {int LEN=G[I].N;        G[i].p[len] = g[i].p[0];                for (int k=0;k<len;k++) if (S21_inst (g[i].p[k],g[i].p[k+1],a,b)) {ans++;   Break         }} return ans;    int main () {int ncases,n;    int ind=1;    scanf ("%d", &ncases);        while (ncases--) {scanf ("%d", &n);            for (int i=0;i<n;i++) {scanf ("%d", &AMP;G[I].N);        for (int k=0;k<g[i].n;k++) scanf ("%d%d", &g[i].p[k].x,&g[i].p[k].y);            } if (n==1) {printf ("Case%d:1\n", ind++);        Continue        } int mmax=0;                    for (int i=0;i<n;i++) for (int. k=0;k<g[i].n;k++) for (int j=i+1;j<n;j++)                        for (int l=0;l<g[j].n;l++) {int ans=solve (g[i].p[k],g[j].p[l],n);                    if (ans > Mmax) mmax = ans;    } printf ("Case%d:%d\n", Ind++,mmax); } return 0;}

HDU3952 Fruit Ninja (geometry)