题目大意：
单纯形*2。。。
#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define EPS 1e-7#define INF 1e10using namespace std;int n,m;namespace Linear_Programming{double A[1010][10100],b[1010],c[10100],v;void Pivot(int l,int e){int i,j;b[l]/=A[l][e];for(i=1;i<=n;i++)if(i!=e)A[l][i]/=A[l][e];A[l][e]=1/A[l][e];for(i=1;i<=m;i++)if(i!=l&&fabs(A[i][e])>EPS){b[i]-=A[i][e]*b[l];for(j=1;j<=n;j++)if(j!=e)A[i][j]-=A[i][e]*A[l][j];A[i][e]=-A[i][e]*A[l][e];}v+=c[e]*b[l];for(i=1;i<=n;i++)if(i!=e)c[i]-=c[e]*A[l][i];c[e]=-c[e]*A[l][e];}double Simplex(){int i,l,e;while(1){for(i=1;i<=n;i++)if(c[i]>EPS)break;if((e=i)==n+1)return v;double temp=INF;for(i=1;i<=m;i++)if( A[i][e]>EPS && b[i]/A[i][e]<temp )temp=b[i]/A[i][e],l=i;if(temp==INF) return INF;Pivot(l,e);}}}int main(){using namespace Linear_Programming;int i,j,x,y,z;cin>>m>>n;for(i=1;i<=m;i++)scanf("%lf",&b[i]);for(i=1;i<=n;i++){scanf("%d%d%d",&x,&y,&z);for(j=x;j<=y;j++)A[j][i]=1;c[i]=z;}double ans=Simplex();cout<<(long long)(ans+0.5)<<endl;return 0;}
，也就越容易失败，还不如怀揣一颗平常心，
