题意：输入n个向量（x，y），要求选出k个，从（0，0）开始画，使得画出来的折线与x轴围成的面积最大。输出面积的二倍。

思路：01背包，dp[i][j]表示y的坐标为i且已经选了j个点的最大面积，然后10包选当前点和不选当前点的最大值。

code：

#include <bits/stdc++.h>
using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000;
const int M=55;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define ft(i,s,n) for (int i=s;i<=n;i++)
#define frt(i,s,t) for (int i=s;i>=t;i--)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);struct node
{int x,y;bool operator <(const node& b)const{return b.x*y>b.y*x;}
}g[M];
int T,n,k;
int dp[M*M][M];
int main()
{scanf("%d",&T);ft(ca,1,T){scanf("%d %d",&n,&k);ft(i,0,n-1) scanf("%d %d",&g[i].x,&g[i].y);sort(g,g+n);cls(dp,-0x3f);dp[0][0]=0;ft(i,0,n-1) frt(j,M*M-1,0) frt(q,k-1,0){if (dp[j][q]>=0)dp[j+g[i].y][q+1]=max(dp[j+g[i].y][q+1],dp[j][q]+(g[i].y+2*j)*g[i].x);}int ans=0;ft(i,0,M*M-1) ans=max(ans,dp[i][k]);printf("Case %d: %d\n",ca,ans);}
}