分析:就是不断递归寻找靠近边界的最优解
学习博客(必须先看这个):
1:http://www.cnblogs.com/autsky-jadek/p/3959446.html
2:http://blog.csdn.net/u013849646/article/details/51524748
注:这里用的最小乘积生成树的思想,和dp结合
每次找满足条件的最优的点,只不过BZOJ裸题的满足条件是形成一棵树
这个题是大于m,生成树借用最小生成树进行求解最优,大于m用dp进行求解最优
#include#include using namespace std;const int N = 8e2+5;typedef long long LL;struct point{ int x, y; point(int x=0,int y=0):x(x),y(y){} point operator -(const point &rhs)const{ return point(x-rhs.x,y-rhs.y); } point operator +(const point &rhs)const{ return point(x+rhs.x,y+rhs.y); }};LL cross(point a,point b){ return 1ll*a.x*b.y-1ll*a.y*b.x;}int n,m,sum;int a[N],b[N],c[N];point p[N];LL dp[N],f[N],ans;point get(){ p[0]=point(0,0),dp[0]=0; for(int i=1;i<=sum;++i)dp[i]=1LL<<60; for(int i=1;i<=n;++i){ for(int j=sum;j>=a[i];--j){ if(dp[j]>dp[j-a[i]]+f[i]){ dp[j]=dp[j-a[i]]+f[i]; p[j]=p[j-a[i]]+point(b[i],c[i]); } } } int ret=m; LL tot=1LL*p[ret].x*p[ret].y; for(int i=m+1;i<=sum;++i){ if(dp[i] =0)return; solve(A,C);solve(C,B);}int main(){ while(~scanf("%d%d",&n,&m)){ sum=0; for(int i=1;i<=n;++i){ scanf("%d%d%d",&a[i],&b[i],&c[i]); sum+=a[i]; } ans=1LL<<60; for(int i=1;i<=n;++i)f[i]=b[i]; point A=get(); for(int i=1;i<=n;++i)f[i]=c[i]; point B=get(); solve(A,B); printf("%I64d\n",ans); } return 0;}