BZOJ 3112: [Zjoi2013]防守战线

真不是我水博客但是它就是那么简单，和它们一模一样

直接套单纯形法的板子

#include<cstdio>
#include<cmath>
#include<iostream>
#define RI register int
#define CI const int&
using namespace std;
const int N=1005,M=10005; 
const double EPS=1e-8;
int n,m,tp,l,r; double a[N][M];
namespace SM //Simplex Method
{
    int id[N+M];
    inline void pivot(CI l,CI e)
    {
        RI i,j; swap(id[n+l],id[e]); double t=a[l][e];
        for (a[l][e]=1,i=0;i<=n;++i) a[l][i]/=t;
        for (i=0;i<=m;++i) if (i!=l&&fabs(a[i][e])>EPS)
        for (t=a[i][e],a[i][e]=j=0;j<=n;++j) a[i][j]-=t*a[l][j];
    }
    inline void simplex(void)
    {
        for (;;)
        {
            RI i; int l=0,e=0; double mi=1e9;
            for (i=1;i<=n;++i) if (a[0][i]>EPS) { e=i; break; } if (!e) break;
            for (i=1;i<=m;++i) if (a[i][e]>EPS&&a[i][0]/a[i][e]<mi) mi=a[i][0]/a[i][e],l=i;    pivot(l,e);
        } 
    }
};
int main()
{
    RI i,j; for (scanf("%d%d",&m,&n),i=1;i<=m;++i) scanf("%lf",&a[i][0]);
    for (i=1;i<=n;++i) for (scanf("%d%d%lf",&l,&r,&a[0][i]),j=l;j<=r;++j) a[j][i]=1;
    return SM::simplex(),printf("%d",(int)(-a[0][0]+0.5)),0;
}