hdu 2907 Diamond Dealer（凸包）

Diamond Dealer

                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                  Total Submission(s): 651    Accepted Submission(s): 190

Problem Description

Mr. Chou is the atworld diamond dealer. It is important that he knows the value of his (twodimensional) diamonds in order to be a succesful businessman. Mr. Chou is tired of calculating the values by hand and you have to write a program that makes the calculation for him.



Figure 2: Example diamond 

The value of a diamond is determined by smoothness of its surface. This
depends on the amount of faces on the surface, more faces means a smoother surface. If there are dents (marked red in gure 2) in the surface of the diamond, the value of the diamond decreases. Counting the number of dents in the surface (a) and the number of faces on the surface that are not in dents (b), the value of the diamond is determined by the following formula: v = -a * p + b * q. When v is negative, the diamond has no value (ie. zero value).

 

Input

The first line of input consists of the integer number n, the number of test cases;
Then, for each test case:
One line containing:
The cost for a dent in the surface of a diamond (0 <= p <= 100); 
The value of a face in the surface of a diamond (0 <= q <= 100); 
The number of vertices (3 <= n <= 30) used to describe the shape of the diamond.
n lines containing one pair of integers (-1000 <=xi,yi <= 1000) describing the surface of the diamond (x0,y0) - (x1,y1) -.....-(xn-1, yn-1) - (x0 ,y0) in clockwise order.
No combination of three vertices within one diamond will be linearly aligned.

 

Output

For each test case, the output contains one line with one number: the value of the diamond.

 

Sample Input

1
10 5 7
0 10
8 4
10 -7
6 -9
-5 -4
-5 7
-2 6

 

Sample Output

15

 

第一道凸包题，这题题意难懂些，就是求凸面和凹面，凸面价值为q，凹面价值为-p，求最后的总价值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
struct node
{
    int x;
    int y;
    int id;
}point[110],stacks[110];
int top;
int hash[110];
int cross(node a,node b,node c)
{
    return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}
double dis(node a,node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)*1.0+(a.y-b.y)*(a.y-b.y)*1.0);
}
bool cmp(node a,node b)
{
    int m=cross(point[0],a,b);
    if(m==0)
        return dis(point[0],a)<dis(point[0],b)?true:false;
    if(m>0)
        return true;
    else
        return false;
}
void graham_scan(int n)
{
    for(int i=1;i<n;i++)
    {
        if(point[i].y<point[0].y||(point[i].y==point[0].y&&point[i].x<point[0].x))
        {
            node temp;
            temp=point[0];
            point[0]=point[i];
            point[i]=temp;
        }
    }
        sort(point+1,point+n,cmp);
        memset(stacks,0,sizeof(stacks));
        stacks[0]=point[0];
        stacks[1]=point[1];
        top=2;
        for(int i=2;i<n;i++)
        {
            while(top>=2&&cross(stacks[top-2],stacks[top-1],point[i])<0)
            {
            top--;
            }
            stacks[top++]=point[i];
        }
}
int main()
{
    int p,q,n,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&p,&q,&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&point[i].x,&point[i].y);
            point[i].id=i;
        }
        graham_scan(n);//凸包
        int cou=0;
        memset(hash,0,sizeof(hash));
        for(int i=0;i<top;i++)
        {
            hash[stacks[i].id]=1;
        }
        hash[n]=hash[0];
        for(int i=0;i<n;i++)//凹面
        {
            if(hash[i]==1&&hash[i+1]==0)
            {
                cou++;
            }
        }

        int ans=top*q-cou*p-cou*q;
        if(ans>0)
        printf("%d\n",ans);
        else
        printf("0\n");
    }
    return 0;
}