POJ 1113 || HDU 1348: wall（凸包问题）

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下面是POJ上的题；

Wall

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 29121 Accepted: 9746

Description

Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he would not listen to his Architect's proposals to build a beautiful brick wall with a perfect shape and nice tall towers. Instead, he ordered to build the wall around the whole castle using the least amount of stone and labor, but demanded that the wall should not come closer to the castle than a certain distance. If the King finds that the Architect has used more resources to build the wall than it was absolutely necessary to satisfy those requirements, then the Architect will loose his head. Moreover, he demanded Architect to introduce at once a plan of the wall listing the exact amount of resources that are needed to build the wall.

Your task is to help poor Architect to save his head, by writing a program that will find the minimum possible length of the wall that he could build around the castle to satisfy King's requirements.

The task is somewhat simplified by the fact, that the King's castle has a polygonal shape and is situated on a flat ground. The Architect has already established a Cartesian coordinate system and has precisely measured the coordinates of all castle's vertices in feet.

Input

The first line of the input file contains two integer numbers N and L separated by a space. N (3 <= N <= 1000) is the number of vertices in the King's castle, and L (1 <= L <= 1000) is the minimal number of feet that King allows for the wall to come close to the castle.

Next N lines describe coordinates of castle's vertices in a clockwise order. Each line contains two integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that represent the coordinates of ith vertex. All vertices are different and the sides of the castle do not intersect anywhere except for vertices.

Output

Write to the output file the single number that represents the minimal possible length of the wall in feet that could be built around the castle to satisfy King's requirements. You must present the integer number of feet to the King, because the floating numbers are not invented yet. However, you must round the result in such a way, that it is accurate to 8 inches (1 foot is equal to 12 inches), since the King will not tolerate larger error in the estimates.

Sample Input

9 100200 400300 400300 300400 300400 400500 400500 200350 200200 200

Sample Output

1628

Hint

结果四舍五入就可以了

题意大致就是要你求将所有点包起来的那个面的最小周长， 以及还有一个以L为半径圆的周长。。

用的是Andrew算法

</pre><pre name="code" class="cpp">#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<sstream>#include<cmath>using namespace std;#define f1(i, n) for(int i=0; i<n; i++)#define f2(i, m) for(int i=1; i<=m; i++)#define f3(i, n) for(int i=n; i>=0; i--)#define M 1005#define PI 3.1415926struct Point{    double x, y;};void sort(Point *p, int n)   //按照x从小到大排序（如果x相同， 按照y从小到大排序）{    Point temp;    f1(i, n-1)    f1(j, n-i-1)    {        if( (p[j].x > p[j+1].x) || (p[j].x==p[j+1].x && p[j].y>p[j+1].y) )        {            temp = p[j];            p[j] = p[j+1];            p[j+1] = temp;        }    }}int cross(int x1, int y1, int x2, int y2)      //看P[i]是否是在其内部。。</span></span>{    if(x1*y2-x2*y1<=0)                        //叉积小于0，说明p[i]在当前前进方向的右边，因此需要从凸包中删除c[m-1],c[m-2]</span><span>          return 0;    else        return 1;}double dis(Point a, Point b)//求两个凸包点之间的长度。。</span><span> {    return sqrt( (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) );}int convexhull(Point *p, Point *c, int n){    int m = 0;    f1(i, n)//下凸包</span><span>      {        while( m>1 && !cross(c[m-2].x-c[m-1].x, c[m-2].y-c[m-1].y, c[m-2].x-p[i].x, c[m-2].y-p[i].y) )            m--;        c[m++] = p[i];    }    int k = m;    f3(i, n-2)//求上凸包</span><span>     {        while( m>k && !cross(c[m-2].x-c[m-1].x, c[m-2].y-c[m-1].y, c[m-2].x-p[i].x, c[m-2].y-p[i].y) )            m--;        c[m++] = p[i];    }    if(n>1)        m--;    return m;}int main(){    Point a[M], p[M];    double sum;    int n, r;    while( cin>>n>>r )    {        sum=0.0;        f1(i, n)        scanf("%lf %lf", &a[i].x, &a[i].y);        sort (a, n);        int m = convexhull(a, p, n);        f2(i, m)        sum+=dis( p[i], p[i-1] );        sum+=2*PI*r;        printf("%.lf\n", sum);    }      return 0;}

我也不知道为什么。。我用lf用G++提交就WA， 用c++就AC。。看讨论区里也说用lf提交错。。把其改为f就对了。。可能G++的输出默认为f把。。。~~(╯﹏╰)b

下面是HDU上AC的代码。。之所以贴出来， 是因为PE过一次。。要注意一下格式。。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<sstream>#include<cmath>using namespace std;#define f1(i, n) for(int i=0; i<n; i++)#define f2(i, m) for(int i=1; i<=m; i++)#define f3(i, n) for(int i=n; i>=0; i--)#define M 1005#define PI 3.1415926struct Point{    double x, y;};void sort(Point *p, int n){    Point temp;    f1(i, n-1)    f1(j, n-i-1)    {        if( (p[j].x > p[j+1].x) || (p[j].x==p[j+1].x && p[j].y>p[j+1].y) )        {            temp = p[j];            p[j] = p[j+1];            p[j+1] = temp;        }    }}int cross(int x1, int y1, int x2, int y2){    if(x1*y2-x2*y1<=0)        return 0;    else        return 1;}double dis(Point a, Point b){    return sqrt( (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) );}int convexhull(Point *p, Point *c, int n){    int m = 0;    f1(i, n)    {        while( m>1 && !cross(c[m-2].x-c[m-1].x, c[m-2].y-c[m-1].y, c[m-2].x-p[i].x, c[m-2].y-p[i].y) )            m--;        c[m++] = p[i];    }    int k = m;    f3(i, n-2)    {        while( m>k && !cross(c[m-2].x-c[m-1].x, c[m-2].y-c[m-1].y, c[m-2].x-p[i].x, c[m-2].y-p[i].y) )            m--;        c[m++] = p[i];    }    if(n>1)        m--;    return m;}int main(){    Point a[M], p[M];    double sum;    int t;    while( cin>>t )    {        while( t-- )        {            sum=0.0;            int n, r;            cin>>n>>r;            f1(i, n)            scanf("%lf %lf", &a[i].x, &a[i].y);            sort (a, n);            int m = convexhull(a, p, n);            f2(i, m)            sum+=dis( p[i], p[i-1] );            sum+=2*PI*r;            printf("%.lf\n", sum);            if(t)                printf("\n");        }    }    return 0;}