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【bzoj3265】志愿者招募加强版 单纯形

2018-01-12 10:35 274 查看
Description



Input



Output



Sample Input

3 3

2 3 4

1 1 2 2

1 2 3 5

1 3 3 2

Sample Output

14

HINT



题解

单纯形裸题

代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
#define mod 10000007
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
#define inf 10000000000.0
#define eps (1e-9)
double ans,a[10005][1005],c[1005],b[10005];
int n,m;
inline void pivot(int l,int e)
{
b[l]/=a[l][e];
for (int i=1;i<=n;i++) if (i!=e) a[l][i]/=a[l][e];
a[l][e]=1/a[l][e];
for (int i=1;i<=m;i++)if (i!=l&&fabs(a[i][e])>eps)
{
b[i]-=a[i][e]*b[l];
for (int 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];
}
ans+=c[e]*b[l];
for (int i=1;i<=n;i++) if (i!=e) c[i]-=c[e]*a[l][i
4000
];
c[e]=-c[e]*a[l][e];
}
inline void simplex()
{
int l,e;
while (1)
{
for (e=1;e<=n;e++) if (c[e]>eps) break;
if (e>n) break;
double t=inf;
for (int i=1;i<=m;i++)if (a[i][e]>eps&&b[i]/a[i][e]<t)
{
t=b[i]/a[i][e];
l=i;
}
pivot(l,e);
}
}
int main()
{
n=read();m=read();
for (int i=1;i<=n;i++) c[i]=read();
for (int i=1;i<=m;i++)
{
int k=read();
while (k--)
{
int s=read(),t=read();
for (int j=s;j<=t;j++) a[i][j]++;
}
b[i]=read();
}
simplex();
printf("%.0lf",ans+eps);
return 0;
}

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