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[POJ 3348] Cows 凸包模板

2015-08-02 12:01 459 查看
http://poj.org/problem?id=3348

题意:用树围城一块地, 求最大的面积。

思路:凸包加多边形面积

点集排序(0ms):

[code]#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 10015;

struct Point{
    int x, y;
    friend bool operator < (Point a, Point b){
        return a.y < b.y || (a.y == b.y && a.x < b.x);
    }
};

//叉积
double mult(Point a, Point b, Point c){
    return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}

int graham(int n, Point *point, Point *res)
{
    sort(point, point+n); //点集排序
    if(n == 0) return 0; res[0] = point[0];
    if(n == 1) return 1; res[1] = point[1];
    if(n == 2) return 2; res[2] = point[2];
    int top = 1;
    //求半个凸包
    for(int i = 2; i < n; i++){
        while(top && mult(point[i], res[top], res[top-1]) >= 0){
            top--;
        }
        res[++top] = point[i];
    }
    //另外半个凸包
    int len = top;
    for(int i = n-2; i >= 0; i--){
        while(top > len && mult(point[i], res[top], res[top-1]) >= 0){
            top--;
        }
        res[++top] = point[i];
    }
    return top;
}

//多边形面积
double Area(int n, Point *res)
{
    double sum = 0.0;
    if(n < 3) return sum;
    for(int i = 0; i < n; i++){
        sum += mult(res[0], res[i], res[(i+1)%n]); //叉积求面积
    }
    return sum / 2.0;
}

Point res[maxn];
Point point[maxn];

int main()
{
    int n;
    while(~scanf("%d", &n)){
        for(int i = 0; i < n; i++){
            scanf("%d%d", &point[i].x, &point[i].y);
        }
        int top = graham(n, point, res);
        double sum = Area(top, res) / 50;
        printf("%d\n",(int)abs(sum));
    }
    return 0;
}


极角排序(16ms):

[code]#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 10015;
const double eps = 1e-8;

struct Point{
    double x, y;
    friend bool operator < (Point a, Point b){
        return a.y < b.y || (a.y == b.y && a.x < b.x);
    }
};

Point res[maxn];
Point point[maxn];

//周长
double length(Point a, Point b){
    return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}

//叉积
double mult(Point a, Point b, Point c){
    return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}

//求面积
double Area(int n, Point *res)
{
    double sum = 0.0;
    if(n < 2) return sum;
    for(int i = 1; i < n; i++){
        sum += fabs(mult(res[0], res[i], res[i+1])); //叉积的绝对值就是构成的三角形面积的两倍
    }
    return sum / 2.0;
}

//极角排序
bool cmp(Point a, Point b)
{
    int k = mult(point[0], a, b);
    return k > 0 || (k == 0 && length(point[0], a) < length(point[0], b));
}

int graham(int n, Point *point, Point *res)
{
    sort(point, point+n, cmp);  //极角排序
    if(n == 0) return 0; res[0] = point[0];
    if(n == 1) return 1; res[1] = point[1];
    if(n == 2) return 2; res[2] = point[2];
    int top = 1;
    for(int i = 2; i < n; i++){
        while(top && mult(point[i], res[top], res[top-1]) >= 0){
            top--;
        }
        res[++top] = point[i];
    }
    return top;
}

int main()
{
    int n;
    while(~scanf("%d", &n)){
        int tx = 100000, ty = 100000, mid;
        for(int i = 0; i < n; i++){
            scanf("%lf%lf", &point[i].x, &point[i].y);
            if(point[i].y < ty || (point[i].y == ty && point[i].x < tx)){
                mid = i;
                tx = point[i].x;
                ty = point[i].y;
            }
        }
        point[mid] = point[0]; //  极角排序要求point[0]是最左下角的点
        point[0].x = tx, point[0].y = ty;
        int top = graham(n, point, res);
        double sum = Area(top, res) / 50;
        printf("%d\n",(int)sum);
    }
    return 0;
}
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