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[1010][10010];
int b[10010],n,m;

il int gi(){
    RG int x=0,q=1; RG char ch=getchar(); while ((ch<'0' || ch>'9') && ch!='-') ch=getchar();
    if (ch=='-') q=-1,ch=getchar(); while (ch>='0' && ch<='9') x=x*10+ch-48,ch=getchar(); return q*x;
}

il void pivot(RG int l,RG int e){
    RG double k=a[l][e]; a[l][e]=1;
    for (RG int i=0;i<=n;++i) a[l][i]/=k; RG int len=0;
    for (RG int i=0;i<=n;++i) if (fabs(a[l][i])>eps) b[++len]=i;
    for (RG int i=0;i<=m;++i)
    if (i!=l && fabs(a[i][e])>eps){
        k=a[i][e],a[i][e]=0;
        for (RG int j=1;j<=len;++j) a[i][b[j]]-=k*a[l][b[j]];
    }
    return;
}

il double simplex(){
    while (1){
    RG int l,e; for (e=1;e<=n;++e) if (a[0][e]>eps) break;
    if (e==n+1) return -a[0][0]; RG double tmp=inf;
    for (RG int i=1;i<=m;++i)
        if (a[i][e]>eps && tmp>a[i][0]/a[i][e]) tmp=a[i][0]/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=1;i<=m;++i) a[i][0]=gi();
    for (RG int i=1;i<=n;++i){
    l=gi(),r=gi(),d=gi(); a[0][i]=d;
    for (RG int j=l;j<=r;++j) a[j][i]=1;
    }
    printf("%lld
",(ll)(simplex()+0.5)); return;
}

int main(){
    work();
    return 0;
}