2011-07-18 16:14:16Morris

Convex Hull (凸包) (By monotone chain (單調鏈))

參照 : 演算法筆記 Convex Hull 的代碼


/**********************************************************************************/
/*  Problem: d919 "最大面積" from                                             */
/*  Language: C                                                                   */
/*  Result: AC (20ms, 590KB) on ZeroJudge                                         */
/*  Author: morris1028 at 2011-07-18 16:07:34                                     */
/**********************************************************************************/


#include<stdio.h>
typedef struct Point {
    int x, y;
}Point;
Point P[100000], CH[100000*2], X[100000];
void MergeSort(int, int);
void Merge(int, int, int);
int cross(Point o, Point a, Point b) {
    return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x);
}
void Calc_Area(int m) {
    int a, sum = 0;
    for(a = 0; a < m-1; a++)
        sum += (CH[a].x*CH[a+1].y - CH[a].y*CH[a+1].x);
    if(sum&1)    printf("%.1lf", sum/2.0);
    else    printf("%d\n", sum/2);
}
void monotone_chain(int N) {
    MergeSort(0, N-1);
    int m = 0, a, t;
    for(a = 0; a < N; a++) {
        while(m >= 2 && cross(CH[m-2], CH[m-1], P[a]) <= 0)
            m--;
        CH[m++] = P[a];
    }
    for(a = N-1, t = m+1; a >= 0; a--) {
        while(m >= t && cross(CH[m-2], CH[m-1], P[a]) <= 0)
            m--;
        CH[m++] = P[a];
    }
    Calc_Area(m);
    m--;
    return;
}
main() {
    int N, a;
    while(scanf("%d", &N) == 1) {
        for(a = 0; a < N; a++)
            scanf("%d %d", &P[a].x, &P[a].y);
        monotone_chain(N);
    }
    return 0;
}
void MergeSort(int l, int h) {
    if(l < h) {
        int m = (l+h)/2;
        MergeSort(l, m);
        MergeSort(m+1, h);
        Merge(l, m, h);
    }
}
void Merge(int l, int m, int h) {
    int In1 = l, In2 = m+1;
    int a, b, Top = 0;
    while(In1 <= m && In2 <= h) {
        if(P[In1].x < P[In2].x || (P[In1].x == P[In2].x && P[In1].y < P[In2].y))
            X[Top++] = P[In1++];
        else
            X[Top++] = P[In2++];
    }
    while(In1 <= m) X[Top++] = P[In1++];
    while(In2 <= h)    X[Top++] = P[In2++];
    for(a = 0, b = l; a < Top; a++, b++)
        P[b] = X[a];
}