HUOJ-10857 最大的面积 凸包+DP

　　题目链接：http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=10857&courseid=55

　　比赛的时候把题目看成取恰好K个点了，，，悲剧。。然后按照正确的题意的话，是比较好做的，求个凸包，然后DP就可以了，f[i][j][k]表示第 i 个点到第 j 点选择k个点的多边形的最大面积，那么f[i][j][k]=Max{ f[i][j][k], f[i][y][k-1]+area(p[i],p[y],p[j]) }就可以了。。

　　这题相当悲剧，题目的数据范围描述错了，k应该是小于等于30，因为题目sb，看了一晚上的代码= =！

//STATUS:C++_AC_0MS_1284KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=60;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e60;
const int dx[4]= {-1,0,1,0};
const int dy[4]= {0,1,0,-1};
const int day[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
//End

struct point{
    double x, y;
}p[N],res[N];

double f[N][N];
int T,n,k;

bool mult(point sp, point ep, point op)
{
    return (sp.x - op.x) * (ep.y - op.y)>= (ep.x - op.x) * (sp.y - op.y);
}

bool operator < (const point &l, const point &r)
{
    return l.y < r.y || (l.y == r.y && l.x < r.x);
}

int graham(point pnt[], int n, point res[])
{
    int i, len, k = 0, top = 1;
    sort(pnt, pnt + n);
    if (n == 0) return 0;
    res[0] = pnt[0];
    if (n == 1) return 1;
    res[1] = pnt[1];
    if (n == 2) return 2;
    res[2] = pnt[2];
    for (i = 2; i < n; i++){
        while (top && mult(pnt[i], res[top], res[top-1]))top--;
        res[++top] = pnt[i];
    }
    len = top;
    res[++top] = pnt[n - 2];
    for (i = n - 3; i >= 0; i--){
        while (top!=len && mult(pnt[i], res[top], res[top-1])) top--;
        res[++top] = pnt[i];
    }
    return top; // 返回凸包中点的个数
}

double arear(point& a,point& b,point& c)
{
    double ret=0;
    ret+=a.x*b.y-a.y*b.x;
    ret+=b.x*c.y-b.y*c.x;
    ret+=c.x*a.y-c.y*a.x;
    return ret/2;
}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,x,y,cnt;
    double ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        for(i=0; i<n; i++){
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        cnt=graham(p,n,res);
        if(cnt<=2 || k<=2){
            printf("0.00
");
            continue;
        }
        if(cnt<=k){
            double sum=0;
            for(i=0;i<cnt;i++)
                sum+=res[i].x*res[(i+1)%cnt].y-res[i].y*res[(i+1)%cnt].x;
            printf("%.2lf
",sum/=2);
            continue;
        }
        ans=0;
        int m=cnt;
        while(m--){
            mem(f,0);
            point t=res[0];
            for(j=0;j<cnt-1;j++)res[j]=res[j+1];
            res[j]=t;
            for(j=2;j<cnt;j++){
                for(x=3;x<=j+1 && x<=k;x++){
                    for(y=x-2;y<j;y++){
                        f[j][x]=max(f[j][x],f[y][x-1]+arear(res[0],res[y],res[j]));
                    }
                }
                ans=max(ans,f[j][k]);
            }
        }

        printf("%.2lf
",ans);
    }
    return 0;
}