bzoj3112 [Zjoi2013]防守战线

正解：线性规划。

直接套单纯形的板子，因为所约束条件都是>=号，且目标函数为最小值，所以考虑对偶转换，转置一下原矩阵就好了。

 //It is made by wfj_2048~
 #include <algorithm>
 #include <iostream>
 #include <complex>
 #include <cstring>
 #include <cstdlib>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <set>
 #define inf (1e15)
 #define eps (1e-12)
 #define il inline
 #define RG register
 #define ll long long

 using namespace std;

 double a[][];
 int b[],n,m;

 il int gi(){
     RG int x=,q=; RG char ch=getchar(); while ((ch<'' || ch>'') && ch!='-') ch=getchar();
     if (ch=='-') q=-,ch=getchar(); while (ch>='' && ch<='') x=x*+ch-,ch=getchar(); return q*x;
 }

 il void pivot(RG int l,RG int e){
     RG double k=a[l][e]; a[l][e]=;
     for (RG int i=;i<=n;++i) a[l][i]/=k; RG int len=;
     for (RG int i=;i<=n;++i) if (fabs(a[l][i])>eps) b[++len]=i;
     for (RG int i=;i<=m;++i)
     if (i!=l && fabs(a[i][e])>eps){
         k=a[i][e],a[i][e]=;
         for (RG int j=;j<=len;++j) a[i][b[j]]-=k*a[l][b[j]];
     }
     return;
 }

 il double simplex(){
     while (){
     RG int l,e; for (e=;e<=n;++e) if (a[][e]>eps) break;
     if (e==n+) return -a[][]; RG double tmp=inf;
     for (RG int i=;i<=m;++i)
         if (a[i][e]>eps && tmp>a[i][]/a[i][e]) tmp=a[i][]/a[i][e],l=i;
     if (tmp==inf) return inf; pivot(l,e);
     }
 }

 il void work(){
     m=gi(),n=gi(); RG int l,r,d;
     for (RG int i=;i<=m;++i) a[i][]=gi();
     for (RG int i=;i<=n;++i){
     l=gi(),r=gi(),d=gi(); a[][i]=d;
     for (RG int j=l;j<=r;++j) a[j][i]=;
     }
     printf("%lld\n",(ll)(simplex()+0.5)); return;
 }

 int main(){
     work();
     return ;
 }